Important Questions for CBSE Class 10 Maths Chapter 10 - Circles

1. There are $\text{3}$ villages A, B and C such that the distance from A to B is $7$km, from B to C is $5$ km and from C to A is $8$ km. The gram pradhan wants to dig a well in such a way that the distance from each village is equal. What should be the location of well? Which value is depicted by gram pradhan?

Ans: 

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 From the given distances, it is clear that A, B, C are non-linear. Hence, A, B, C will form a triangle and lie on the circumference of the circle formed by the vertices of the triangle. 

The point equidistant from the points will be the circumcentre. Hence, the location of the well will be at the centre of the circle. 

The values depicted by gram Pradhan are social, honesty and equality.

2. People of village want to construct a road nearest to a circular village Rampur. The road cannot pass through the village. But the people want that road should be at the shortest distance from the center of the village

(i) which road will be the nearest to the center of village?

Ans: The road passing through the tangent of the circle will be nearest to the centre of the village.

(ii) which value is depicted by the people of village?

Ans: Economical 

3. Four roads have to be constructed by touching village Khanpur in circular shape of radius $1700$ m in the following manner. 

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Savita got contract to construct the roads $\mathbf{AB}$ and $\mathbf{CD}$ while Vijay got contract to construct $\mathbf{AD}$ and $\mathbf{BC}$. Prove that $\mathbf{AB+CD=AD=BC}$ . Which value is depicted by the contractor?

Ans: Mark points E, F, G and H on the circle as shown in the figure,

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Now,

AH = AE , BE = BF , CF = CG , DG = AH …(Tangents meeting outside the circle at a point are always equal)

The value depicted by the constructor is gender equality.

4. Two roads starting from P are touching a circular path at $A$ and $B$ . Sarita ran from $P$ to $A$ $10$ km and Ramesh ran from$P$ to $B$ .

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(i) If Sarita wins the race than how much distance Ramesh ran?

Ans: We know that, PA = PB , PB = 10 

Hence, Ramesh ran $10$ km.

(ii) Which value is depicted?

Ans: The value depicted is gender equality.

5. A farmer wants to divide a sugarcane of $7$ ft length between his son and daughter equally. Divide it geometrically, considering sugarcane as a line of $7$ cm, using construction.

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(i) Find the length of each part.

Ans: 

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The length of each half is $3.5$ft.

(ii) Which value is depicted?

Ans: The value depicted is gender equality.

1 Marks Questions

1. How many tangents can a circle have?

Ans: A circle can have infinitely many tangents since there are infinitely many points on the circumference of the circle and at each point of it, it has a unique tangent.

2. The perimeter of a sector of a circle of radius $8$cm is $25$m, what is area of sector?

1. $50c{{m}^{2}}$ 

2. $42c{{m}^{2}}$ 

3. $52c{{m}^{2}}$

4. none of these

Ans: Given radius= 8 cm and perimeter of sector=25 cm

$ \text{Perimeter of a sector of circle=}\left( \dfrac{\theta }{{{360}^{\circ }}}\times 2\pi r \right)+2r $

$ \Rightarrow 25=\left[ \dfrac{\theta }{{{360}^{\circ }}}\times 2\pi \left( 8 \right) \right]+2\left( 8 \right) $

$ \Rightarrow 25=\dfrac{\theta }{{{360}^{\circ }}}\pi \times 16+16 $

$ \Rightarrow 25-16=\dfrac{\theta }{{{360}^{\circ }}}\pi \times 16 $ 

$ \Rightarrow \dfrac{9}{16}=\dfrac{\theta }{{{360}^{\circ }}}\pi $

$ \text{Area of a sector of a circle=}\dfrac{\theta }{{{360}^{\circ }}}\times \pi {{r}^{2}} $

$ =\left( \dfrac{\theta }{{{360}^{\circ }}}\pi \right)\times {{r}^{2}} $

$ =\dfrac{9}{16}\times {{\left( 8 \right)}^{2}} $

$ =\dfrac{9}{16}\times 64 $

$ =36 $

Hence, (d) none of these.

3. In figure given below $PA$ and $PB$ are tangents to the circle drawn from an external point $P$ . $CD$ is a third tangent touching the circle at $Q$ . If $PA=10cm$ and $DQ=2cm$ . What is length of $PC$?

1. 8cm 

2. 7cm 

3. 4cm 

4. none of these

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Ans: (a) 8 cm

4. Tangent of circle intersect the circle

1. Only one point

2. Two points

3. Three points

4. None of these

Ans: (a) Only one point

5. From a point $Q$ , the length of the tangent to a circle is $24cm$ and the distance of $Q$ from the centre is $25cm$ The radius of the circle is

1. 7cm

2. 12cm 

3. 15cm 

4. 24.5cm 

Ans: (a) $7cm$ 

6. How many tangents can a circle have?

(a) 1

(b) 2 

(c) 0 

(d) infinite

Ans: (d) infinite

7. If $PA$ and $PB$ are tangents from a point $P$ lying outside the circle such that $PA=10cm$ and $ \angle(APB) = {60^\circ}$. Find length of chord$AB$.

1. 10cm 

2. 20cm 

3. 30cm 

4. 40cm 

Ans: (a) $10cm$ 

8. A tangent $PQ$ at a point $P$ to a circle of radius $5cm$ meets a line through the centre $O$ at a point $Q$ , so that $OQ=13cm$ , then length of $PQ$ is

1. 11cm

2. 12cm 

3. 10cm 

4. None of these

Ans: (b) $12cm$ 

9. If tangents $PA$ and $PB$ from a point $P$ to a circle with centre $O$ are inclined to each other at angle of ${{80}^{\circ }}$, then is equal to

1. \[\mathbf{{{50}^{\circ }}}\] 

2. $\mathbf{{{60}^{\circ }}}$ 

3. $\mathbf{{{70}^{\circ }}}$ 

4. $\mathbf{{{80}^{\circ }}}$ 

Ans: (a) ${{50}^{\circ }}$ 

10. How many tangents can a circle have?

(a) 1

(b) 2 

(c) 0 

(d) infinite

Ans: (d) infinite

11. If $PA$ and $PB$ are tangents from a point $P$ lying outside the circle such that $PA=10cm$ and $\angle{APB} = {{60}^\circ}$. Find length of chord $AB$.

(a) 10cm 

(b) 20cm 

(c) 30cm 

(d) 40cm 

Ans: (a) $10cm$ 

12. The length of tangent drawn to a circle with radius $3cm$ from a point $5cm$ from the centre of the circle is

(a) 6cm 

(b) 8cm 

(c) 4cm 

(d) 7cm 

Ans: (c) $4cm$ 

13. A circle touches all the four sides of a quadrilateral $ABCD$ whose sides $AB=6cm$, $BC=7cm$, $CD=4cm$ then$AD=$ ____

(a) 2cm 

(b) 3cm 

(c) 5cm 

(d) 6cm 

Ans: (b)$3cm$ 

14. If a point lies on a circle, then what will be the number of tangents drawn from that point to the circle?

(a) 1 

(b) 2 

(c) 3 

(d) infinite

Ans: (a) $1$ 

15. A quadrilateral $ABCD$ is drawn to circumscribe a circle if $AB=4cm$, $CD=7cm$, $BC=3cm$ , then length of $AD$ is

1. 7cm

2. 2cm 

3. 8cm 

4. none of these

Ans: (c) $8cm$ 

16. A tangent $PQ$ at point $P$ of a circle of radius $12cm$ meets a line through the centre $O$ to a point $Q$ so that $OQ=20cm$, then length of $PQ$ is

(a) 14cm 

(b) 15cm 

(c) 16cm 

(d) 10cm 

Ans: (d) $10cm$ 

17. A line intersecting a circle in two points is called

1. Tangent

2. Secant

3. Diameter

4. None of these

Ans: (b) secant

18. The length of tangent from a point $A$ at a distance of $5cm$ from the centre of the circle is $4cm$. What will be the radius of circle?

1. 1cm

2. 2cm

3. 3cm 

4. none of these

Ans: (c) $3cm$ 

19. In the figure given below, $PA$ and $PB$ are tangents to the circle drawn from an external point $P$ . $CD$ is a third tangent touching the circle at $Q$. If $PB=12cm$ and $CQ=3cm$, what is the length of$PC$?

1. 9cm

2. 10cm 

3. 1cm 

4. 13cm 

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Ans: (a) $9cm$ 

20. The tangent of a circle makes angle with radius at point of contact

1. $\mathbf{{{60}^{\circ}}}$ 

2. $\mathbf{{{30}^{\circ}}}$ 

3. $\mathbf{{{90}^{\circ}}}$ 

4. none of these

Ans: (c) ${{90}^{\circ}}$ 

21. If tangent $PA$ and$PB$ from a point $P$ to a circle with centre $0$ are inclined to each other at an angle of ${{80}^{\circ}}$ , then what is the value of

1. $\mathbf{{{30}^{\circ}}}$ 

2. $\mathbf{{{50}^{\circ}}}$ 

3. $\mathbf{{{70}^{\circ}}}$ 

4. $\mathbf{{{90}^{\circ}}}$ 

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Ans: (b) ${{50}^{\circ}}$ 

2 Marks Questions

1. Fill in the blanks:

(i) A tangent to a circle intersects it in _______________ point(s).

Ans: one 

(ii) A line intersecting a circle in two points is called a _______________.

Ans: secant

(iii) A circle can have _______________ parallel tangents at the most.

Ans: two

(iv) The common point of a tangent to a circle and the circle is called _______________.

Ans: point of contact.

2. A tangent $PQ$ at a point $P$ of a circle of radius $5cm$ meets a line through the centre $O$ at a point $Q$ so that $OQ=12cm$ . Length $PQ$ is:

(A) 12cm 

(B) 13cm 

(C) 8.5cm

(D) $\mathbf{\sqrt{119}cm}$ 

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Ans: (D) $\because$ $PQ$ is the tangent and $OP$ is the radius through the point of contact. $\angle{OPQ} ={{90}^\circ}$ … (The tangent at any point of a circle is to the radius through the point of contact)

In right triangle $OPQ$ ,

$O{{Q}^{2}}=O{{P}^{2}}+P{{Q}^{2}}$ …(By Pythagoras theorem)

$ \Rightarrow \left( {{12}^{2}} \right)\left( {{5}^{2}} \right)+P{{Q}^{2}} $

$ \Rightarrow 144=25+P{{Q}^{2}} $

$ \Rightarrow P{{Q}^{2}}=144-25=119 $

$ \Rightarrow PQ=\sqrt{119} $

3. Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.

Ans: 

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4. From a point $Q$, the length of the tangent to a circle is $24cm$ and the distance of $Q$ from the centre is $25cm$ . The radius of the circle is:

1. 7cm 

2. 12cm 

3. 15cm 

4. 24.5cm 

Ans:

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$ \angle{OPQ} ={{90}^\circ}$ 

(The tangent at any point of a circle is $\bot $ to the radius through the point of contact)

In right triangle $OPQ$,

$O{{Q}^{2}}=O{{P}^{2}}+P{{Q}^{2}}$ …(By Pythagoras theorem)

$ \Rightarrow ({{25}^{2}}) = (O{{P}^{2}}) + ({{24}^{2}}) $

$ \Rightarrow 625= (O{{P}^{2}})+576 $

$ \Rightarrow O{{P}^{2}}=625-576=49 $

$ \Rightarrow OP=7cm $

5. In figure, if $TP$ and $TQ$ are the two tangents to a circle with centre $O$ so that $\angle{POQ} = {{110}^{\circ}}$, then $\angle{PTQ} =$ 

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1. $\mathbf{{{60}^{\circ}}}$ 

2. $\mathbf{{{70}^{\circ}}}$ 

3. $\mathbf{{{80}^{\circ}}}$ 

4. $\mathbf{{{90}^{\circ}}}$ 

Ans: (B) $\angle{POQ} = {{110}^\circ}$ $\angle{OPT} ={{90}^\circ}$ and ${\angle{OQT}} = {{90}^\circ}$ …(The tangent at any point of a circle is to the radius through the point of contact)

In quadrilateral $OPTQ$,

$ \angle{POQ} + \angle{OPT} + \angle{OQT} + \angle{PTQ} = {360^{\circ}}$

(Angle sum property of quadrilateral)

$ \Rightarrow {{110}^{\circ}} + {{90}^\circ} + {{90}^\circ} + {\angle{PTQ}} = {360^{\circ}}$

$ \Rightarrow {{290}^\circ} + {\angle{PTQ}} = {{360}^\circ} $

$ \Rightarrow \angle{PTQ} ={{70}^\circ}$

6. If tangents $PA$ and $PB$ from a point $P$ to a circle with centre $O$ are inclined to each other at angle of ${{80}^\circ}$, then $\angle{POA}$ is equal to:

(A) $\mathbf{{{50}^\circ}}$

(B) $\mathbf{{{60}^\circ}}$ 

(C) $\mathbf{{{70}^\circ}}$ 

(D) $\{{80}^\circ}}

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Ans: (A) $\because \angle{OPQ} ={{90}^\circ}$… (The tangent at any point of a circle is perpendicular to the radius through the point of contact.

$\angle{OPA} = {\dfrac{1}{2}} \angle{BPA} $ …(Centre lies on the bisector of the angle between the two tangents)

In $\vartriangle OPA$ ,

$\angle{OAP} +\angle{OPA} + \angle{POA} = {{180}^\circ}$ …(Angle sum property of a triangle)

$ \Rightarrow {{90}^\circ} + {{40}^\circ} + \angle{POA} ={{180}^\circ}$ $\Rightarrow {{130}^\circ} + \angle{POA} ={{180}^\circ}$

$ \Rightarrow \angle{POA} ={{50}^\circ} $

7. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

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Ans: Given: $PQ$ is a diameter of a circle with centre $O$.

The lines $AB$ and$CD$ are the tangents at $P\text{ and Q}$ respectively.

To Prove: $AB\parallel CD$ 

Proof: Since $AB$ is a tangent to the circle at $P$ and $OP$ is the radius through the point of contact.

$\therefore \angle{OPA} ={{90}^\circ}$…(i) (The tangent at any point of a circle is perpendicular to the radius through the point of contact)

$CD$ is a tangent to the circle at $Q$ and $OQ$ is the radius through the point of contact.

$\therefore \angle{OQD} = {{90}^\circ}$…(ii) (The tangent at any point of a circle is perpendicular to the radius through the point of contact)

From eq. (i) and (ii),

$\angle{OPA} = \angle{OQD}$ 

But these form a pair of equal alternate angles also,

$\therefore AB\parallel CD$ 

8. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Ans: We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact and the radius essentially passes through the centre of the circle, therefore the perpendicular at the point of contact to the tangent to a circle passes through the centre.

9. The length of a tangent from a point $A$ at distance $5cm$ from the centre of the circle is $4cm$. Find the radius of the circle.

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Ans: We know that the tangent at any point of a circle is to the radius through the point of contact.

$ \therefore \angle{OPA} ={{90}^\circ}$

$ \therefore O{{A}^{2}} = O{{P}^{2}} + A{{P}^{2}}$ (by pythagoras theorem)

$ \Rightarrow ({{5}^{2}}) = (O{{P}^{2}}) + ({{4}^{2}})$

$ \Rightarrow 25 = (O{{P}^{2}}) + 16$

$ \Rightarrow OP = 3cm$

10. Two concentric circles are of radii $5cm$ and $3cm$. Find the length of the chord of the larger circle which touches the smaller circle.

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Ans: Let $O$ be the common centre of the two concentric circles.

Let $AB$ be a chord of the larger circle which touches the smaller circle at $P$.

Join $OP$ and $OA$ 

Then, $ \angle{OPA} = {{90}^\circ}$ 

(The tangent at any point of a circle is to the radius through the point of contact)

$ \therefore O{{A}^{2}} = O{{P}^{2}} + A{{P}^{2}}$ (by pythagoras theorem)

$ \Rightarrow ({{5}^{2}}) = ( {{3}^{2}}) + (A{{P}^{2}})$

$ \Rightarrow 25 = 9 + A{{P}^{2}}$

$ \Rightarrow A{{P}^{2}} = 16$

$ \Rightarrow AP = 4cm$

Since the perpendicular from the centre of a circle to a chord bisects the chord, therefore

AB = BP = 4cm 

$ \Rightarrow$ AB = AP + BP + AP = 2AP = $2\times 4 = 8cm$ 

11. A quadrilateral $ABCD$ is drawn to circumscribe a circle (see figure). Prove that: $AB+CD=AD+BC$ 

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Ans: We know that the tangents from an external point to a circle are equal.

AP = AS ...(i)

BP = BQ ...(ii)

CR = CQ ...(iii)

DR = DS ...(iv)

On adding eq. (i), (ii), (iii) and (iv), we get

$ (AP+BP) + (CR+DR) = (AS+BQ) + (CQ+DS) $

$ \Rightarrow AB + CD = (AS+DS) + (BQ+CQ) $

$ \Rightarrow AB + CD = AD + BC$

12. In two concentric circles prove that all chords of the outer circle which touch the inner circle are of equal length.

Ans:

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$AB$ and $CD$ are two chords of the circle which touch the inner circle at $M$ and $N$ respectively

$\therefore OM=ON$ 

$\Rightarrow AB=CD$…(AB and CD are two chords of larger circle)

13. $PA$ and $PB$ are tangents from $P$ to the circle with centre $O$ . At the point $M$, a tangent is drawn cutting $PA$ at $K$ and $PB$ at $N$ . Prove that $KN=AK+BN$.

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Ans: We know that the lengths of the tangents drawn form an external point to a circle are equal.

$ \therefore PA=PB.....(i) $

$ KA=KM.....(ii) $

$ NB=NM.....(iii) $

(ii) + (iii) 

$\Rightarrow$ KA + NB = KM + NM 

$ \Rightarrow$ AK + BN = KM + MN 

$ \Rightarrow$ AK + BN = KN 

14. In the given figure, $O$ is the centre of the circle with radius $5cm$ and $AB\parallel CD$. If $AB=6cm$, find $OP$.

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Ans: Given $ OP\bot AB$ 

OP bisects AB

$ \therefore AP=\dfrac{1}{2}AB=\dfrac{1}{2}\times 6=3cm$

$ \vartriangle OAP, O{{A}^{2}} = O{{P}^{2}} + A{{P}^{2}} $

$ \Rightarrow {{5}^{2}} = O{{P}^{2}} + {{3}^{2}}$

$ \Rightarrow OP = 4cm $

15. Prove that the tangents at the end of a chord of a circle make equal angles with the chord.

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Ans: In $\vartriangle ADB$ and $\vartriangle ADC$ 

BD = DC

$ \angle{ADB} = \angle{ADC} = {{90}^\circ}$ 

AD = AD (common) 

$ \therefore \vartriangle \text{ADB} \cong \vartriangle \text{ADC}$ (SAS)

$ \therefore \angle{ABD} = \angle{ACD}$ (by CPCT)

16. Find the locus of the centre of circles which touch a given line at a given point.

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Ans: Let $APB$ be the given line and let a circle with centre $O$ touches the line $AB$ at $P$ . Then, let there be another circle with centre ${O}'$ which touches the line $AB$ at $P$ .

Thus,

$ \because \angle{OPB} = {{90}^{\circ}} $

$ \angle {O}'PB = {{90}^{\circ}}$ 

It is clear that both the centers of the circles lies on the same line that is perpendicular to the given line.

So, the locus of the centers of the circles that touch the given line at a given point is the line perpendicular to the given line passing through the given point.

17. If $PA$ and $PB$ are tangents drawn from external point $P$ such that $PA=10cm$ and, find the length of chord $AB$ 

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Ans: From the properties of the tangents we can say that,

$\because \angle APB={{60}^{\circ}}$ 

$\angle AOB={{120}^\circ}$ (O is centre of circle)

$ \angle{OAB} = \angle{OBA} = {{30}^{\circ}} $

$ \therefore \angle{PAB} = {{60}^{\circ}}, \angle{PBA} = {{60}^{\circ}}$

As all the angles of the triangle PAB are equal,

$ \therefore \Delta {PAB}$ is equilateral triangle

$ \therefore \text{AB = PA = 10cm}$ 

18. If $AB$, $AC$ and $PQ$ are tangents in the given figure and $AB=25cm$, find the perimeter of $\vartriangle APQ$ 

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Ans: Perimeter of $\Delta APQ = AP + AQ + PQ$ 

= AP + AQ + PX + XQ 

As we know that XQ = QC and PX = PB we get,

= (AP+PB) + (AQ+QC)

= AB + AC 

= 2AB = $2\times 25$ = 50cm 

19. In the given figure, find the perimeter of $\Delta ABC$ if $CP=10cm$.

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Ans: From the figure we can see that $BA$ touches the circle at $R$ 

Tangents drawn from external point to the circle are equal.

$ \because AP = AR, BR = BQ$ 

And $CP=CQ$ 

$\therefore$ Perimeter of $\Delta ABC = AB + BC + AC$ 

= (BR + RA) + BC + AC 

= (QB + AP) + BC + AC 

= AP + AC + QB + BC

= PC + CQ

= 2CP 

= $2\times 10$

= 20cm

20. Find the unknown length x. Given PA = 5cm and PB = 8cm.

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Ans: PT is tangent to a circle and PAB is a secant.

$ \therefore PA\times PB = P{{T}^{2}}$

$ \Rightarrow 5(5 + x) = {{8}^{2}} $

$ \Rightarrow 25 + 5x = 64 $

$ \Rightarrow x = \dfrac{39}{8} = 7.8cm$

21. In the given figure, $OD$ is perpendicular to the chord $AB$ of a circle whose centre is O. If BC is a diameter, find $\dfrac{CA}{OD}$.

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Ans: Since $BC$ is a diameter

$ \therefore \angle CAB = {{90}^\circ}$

also OD $\bot$ AB

$ \therefore \angle ODB = {{90}^\circ}$

Considering the triangles $\Delta ACB, \Delta DOB$ we have,

$\angle CAB=\angle ODB = {{90}^\circ}$

$\angle ABC=\angle DBC$

Now, as $DO\parallel AC$ we have,

$\angle ACB=\angle DOB$

By AAA criteria we can say that $\Delta ACB\sim \Delta DOB$ 

So, the sides will be in proportion that is,

$\therefore \dfrac{CA}{OD} = \dfrac{CB}{OB} = \dfrac{2r}{r} = 2$

22. In the given figure, XP and XQ are tangents from X to the circle with centre O. R is a point on the circle such that ARB is a tangent to the circle prove that XA + AR = XB + BR.

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Ans: In the given figure, XP and XQ are tangents from external point

$ \therefore XP=XQ....(i) $

$ AR=AP....(ii) $

$ BR=BQ....(iii) $ (Length of tangents are equal from external point)

XP = XQ 

XP + AP = XB + BQ …(By (ii) and (iii))

XA + AR = XB + BR …(By (ii) and (iii))

23. Prove that the segment joining the points of contact of two parallel tangents, passes through the centre.

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Ans: Given two parallel tangents AB and CD of a circle with centre O touching the circle at P, Q respectively.

Draw line $OR\parallel AB$ from the center O and join PQ then we have,

$\angle{POQ} + \angle{ROQ} = {{180}^\circ}$ 

This proves that the point $O$ lies on $PQ$.

24. In figure, if $OL=5cm$ ,$OA=13cm$ , then length of $AB$ is

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Ans:

$AB = 2AL = \sqrt[2]{O{{A}^{2}}-O{{L}^{2}}}$

$ = \sqrt[2]{{{13}^{2}}-{{5}^{2}}}$

$ = \sqrt[2]{169-25}$

$ = \sqrt[2]{144} $

$ = 2\times 12 $

$ = 24cm $

25. In the given figure, ABCD is a cyclic quadrilateral and PQ is a tangent to the circle at C. If BD is a diameter, and, find $\angle OCQ={{40}^\circ}$ and $\angle ABD={{60}^\circ}$ ,Find $\angle BCP$ 

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Ans: BD is a diameter

$ \therefore \angle BCD={{90}^\circ}$ {angle in the semicircle}

 $ \angle {BCP=18}{{\text{0}}^\circ} - {{90}^\circ} - {{40}^\circ} = {{50}^\circ}$ 

26. Two chords AB and CD of a circle intersect each other at P outside the circle. If AB = 5 cm $AB=5cm$ , $BP=3cm$ and $PD=2cm$ , find $CD$ 

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Ans: Two chords AB and CD of a circle intersect each other at P

PA × PB = PC × PD (length of tangent from P)

$ \Rightarrow (AB+PB) \times PB = (PD+PC)PD$

$ \Rightarrow (5+3)(3) = ( 2 + x) \times 2$ 

$ \Rightarrow 24=(2+x)\times 2$

$ \Rightarrow x = 10 $

$ \Rightarrow CD = 10cm$

27. In the adjoining figure, if AD, AE and BC are tangents to the circle at D, E and F respectively, then prove that 2AD = AB + BC + CA.

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Ans: CD = CF, BE = BF

$ \Rightarrow$ CD + BE = CF + BF = AC + CF

Now AD = AC + CD = AC + CF 

AE = AB + BE = AB + BF 

$ \therefore$ AD + AE = AB + AC + BC 

$ \Rightarrow$ 2AD = AB + BC + AC 

28. In figure, PA and PB are tangents from P to the circle with centre O. R is a point on the circle, prove that PC + CR = PD + DR.

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Ans: Since length of tangents from an external point to a circle are equal in length

$ \therefore$ PA = PB

CA = CR ...(i)

And DB = DR and PA = PB 

PC + CA = PD + DB 

$\Rightarrow$ PC + CR = PD + DR

29. The length of tangents from a point A at distance of 26 cm from the centre of the circle is 10cm, what will be the radius of the circle?

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Ans: Since tangents to a circle is perpendicular to radius through the point of contact

$\therefore \angle OTA = {{90}^\circ}$ 

In right $\vartriangle OTA = {{90}^\circ}$, we have

$ O{{A}^{2}} = O{{T}^{2}} + A{{T}^{2}}$ 

$ \Rightarrow {{\left(26 \right)}^{2}} = O{{T}^{2}} + {{\left( 10 \right)}^{2}}$

$ \Rightarrow O{{T}^{2}} = 676 - 100$

$ \Rightarrow OT = 24 $

30. In the figure, given below PA and PB are tangents to the circle drawn from an external point P. CD is the third tangent touching the circle at Q. If PB = 10 cm and CQ = 2cm, what is the length of PC?

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Ans: PA = PB = 10cm and CQ = CA = 2cm and PC = PA - CA = 10 - 2 = 8cm

3 Marks Questions

1. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

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Ans: $ \angle OPA = {{90}^\circ}....(i)$

$\angle OCA = {{90}^\circ}....(ii)$

(Tangent at any point of a circle is perpendicular to the radius through the point of contact)

OAPB is quadrilateral.

$\therefore \angle APB + \angle AOB + \angle OAP + \angle OBP = {{360}^\circ}$ 

(Angle sum property of a quadrilateral)

 $\Rightarrow \angle APB + \angle AOB + {{90}^\circ} + {{90}^\circ} = {{360}^\circ}$ 

(From eq. (i) and (ii))

$\Rightarrow \angle APB + \angle AOB = {{180}^\circ}$ 

$\therefore \angle APB$ and $\angle AOB$ are supplementary.

2. Prove that the parallelogram circumscribing a circle is a rhombus.

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Ans: Given: ABCD is a parallelogram circumscribing a circle.

To Prove: ABCD is a rhombus.

Proof: Since, the tangents from an external point to a circle are equal.

$ \therefore$ AP = AS ....(i)

BP = BQ ....(ii)

CR = CQ ....(iii)

DR = DS ....(iv)

On adding eq. (i), (ii), (iii) and (iv), we get

(AP+BP) + (CR + DR) = (AS+BQ) + (CQ+DS)

$\Rightarrow$ AB+CD = (AS+DS) + (BQ+CQ)

$ \Rightarrow$ AB+CD = AD+BC

$ \Rightarrow$ AB+AB = AD+AD

$ \Rightarrow$ 2AB = 2AD 

$ \Rightarrow$ AB = AF 

but AB = CD and AD = BC 

$\therefore$ AB = BC = CD = AD 

$ \therefore$ Parallelogram ABCD is a rhombus.

3. Two tangents PA and PB are drawn from an external point P with centre O as shown in figure. If they are inclined to each other at an angle of $100^\circ$, then what is the value of $\angle AOB$?

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Ans: $\therefore PA$ and $PB$ are tangents and $O$ is the centre of the circle

$ \therefore OA \bot PA, OB\bot PB$

$ \therefore \angle PAO + \angle PBO = {{180}^\circ}$

$ \therefore$ Quadrilateral PAOB is cyclic 

$ \therefore \angle{APB} + \angle{AOB} = {{180}^\circ}$

$ \therefore {{100}^\circ} + \angle AOB={{180}{^\circ}}$ 

$ \therefore \angle AOB = {{180}^\circ} - {{100}^\circ} = {{80}^\circ}$

4. Two concentric circles are of radii 5 cm and 3cm, find the length of the chord of the larger circle which touches the smaller circle.

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Ans: $\because PQ$ is the chord of the larger circle which touches the smaller circle at the point M.

Since OM is tangent at the point M to the smaller circle with centre O.

$\therefore OM \bot PQ$ 

$ \because$ PQ is a chord of the bigger circle and OM $\bot$ PQ 

$ \therefore OM\text{ bisects PQ} $

$ \therefore \text{PQ=2PM} $

$ \text{In }\vartriangle \text{OPM, PM=}\sqrt{O{{P}^{2}}-O{{M}^{2}}} $

$ \text{=}\sqrt{{{5}^{2}}-{{3}^{2}}} $

 $ \text{=}\sqrt{25-9}=4 $

$\therefore $ chord $PQ=2PM=8cm$ 

$\therefore $ length of cord $PM=8cm$

6. PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at point T. Find the length TP.

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Ans: Join $OT$ 

TP = PQ [tangents from T upon the circle]

$\therefore OT \bot PQ$ 

And $OT$ bisects $PQ$ 

$\therefore PR = PQ = 4cm$ 

Now,

$ OR = \sqrt{O{{P}^{2}}+P{{R}^{2}}}$

$ OR=\sqrt{{{5}^{2}}-{{4}^{2}}}=3cm $

Now $\angle{TPR}+\angle{PTR}$

$ \therefore \angle{RPO}=\angle{PTR} $

$ \vartriangle TRP\sim \vartriangle TRQ$ (by AA similarity)

$ \therefore \dfrac{TP}{PO}=\dfrac{RP}{RO}$

$ \Rightarrow \dfrac{TP}{5}=\dfrac{4}{3}$

$ \Rightarrow TP=\dfrac{20}{3}cm $

7. A circle is touching the side BC of at P and touching AB and AC produced at Q and R respectively. Prove that AQ = $\dfrac{1}{2}$ (perimeter of $\vartriangle ABC$).

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Ans: We know that the two tangents drawn to a circle from an external point are equal.

$ \therefore$ AQ = AR, BP = BQ, CP = CR

$ \therefore$ Perimeter of $\vartriangle$ ABC = AB + BC + AC 

= AB + BQ + CR + AC $\because$ BP = BQ, PC = CQ

$= AQ + AR = 2AQ = 2AR \because AQ = AR$

= AQ = AR = $\dfrac{1}{2}$ perimeter of $\vartriangle$ ABC

8. If PA and PB are two tangents drawn from a point P to a circle with centre O touching it at A and B. Prove that OP is the perpendicular bisector of AB.

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Ans: Let OP intersect AB at a point C, we have to prove that AC = CB and

$\angle ACP = \angle BCP = {{90}^\circ}$ 

$\because PA,PB$ are two tangents from a point P to the circle with centre O

$\therefore \angle APO = \angle BPO$…(O lies on the bisector of $\angle APB$)

In ACP and BCP we have

AB = BP (because tangents from P to the circle are equal)

PC = PC (Common)

$\angle APO = \angle BPO$ (proved)

$ \therefore \vartriangle ACP \cong \vartriangle BCP$ (By SAS rule)

$ \therefore {AC = CB}$ (CPCT)

And $\angle ACP = \angle BCP$ (CPCT)

But $\angle ACP + \angle BCP = {{180}^\circ}$ 

$\Rightarrow \angle ACP = \angle BCP = {{90}^\circ}$ 

Hence, $OP$ is a perpendicular bisector of $AB$.

9. In the given figure, PQ is tangent at point R of the circle with centre O. If $\angle TRQ = {{30}^\circ}$ , find $\angle PRS$ 

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Ans: Given PQ is tangent at point R and $\angle TRQ = {{30}^\circ}$ 

$ \angle PRQ = {{180}^\circ}$

$ \angle QRT = {{30}^\circ}$

$\angle TRS={{90}^\circ} [\because $ Tangent of a circle is perpendicular to Radius]

$ \therefore \angle PRS = {{180}^\circ}-{{120}^\circ}={{60}^\circ}$ 

$ \therefore \angle PRS={{180}^\circ}-{{120}^\circ}={{60}^\circ}$

10. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

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Ans: Let the circle touch the sides AB, BC, CD and DA at the points P, Q, R, and S respectively.

Join OP, OQ, OR and OS.

Join OA, OB, OC and OD.

Since the two tangents drawn from an external point subtend equal angles at the centre.

$\angle 1=\angle 2,\angle 3=\angle 4,\angle 5=\angle 6,\angle 7=\angle 8$ 

But, $\angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7+\angle 8+={{360}^\circ}$ 

$\therefore (AP+DP)+(BR+CR)$ 

$=AQ+DS+BQ+CS$ 

$=(AQ+BQ)+(CS+DS)$ 

$\Rightarrow AD+BC=AB+BC$ 

$\Rightarrow BC+BC=AB+AB$

$[\because AB=DC,AD=BC]$ 

$\Rightarrow 2BC=2AB$ 

$\Rightarrow BC=AB$ 

Hence, parallelogram ABCD is a rhombus.

11. If two tangents are drawn to a circle from an external point then

(i) they subtend equal angles at the centre.

(ii) they are equally inclined to the segment joining the centre to that point.

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Ans: Given on a circle C, two tangents AP and PB are drawn from an external point P.

To prove:

$ (i)\angle AOP=\angle BOQ $

$ (ii)\angle OPA=\angle OPB $

Construction: Join AO, PO and BO

Proof: In $\Delta PAO\text{ and }\Delta PBO$, 

AP = PB (Length of the tangents drawn from an external point)

AO = BO (Radii of the same circle)

PO = PO (common)

$ \Delta PAO \cong \Delta PBO$ (by SSS theorem of congruence)

$ \text{(i)}\angle {AOP} = \angle {BOQ}$ [CPCT]

$ (ii)\angle OPA = \angle OPB$ [by CPCT]

12. Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that $\angle PTQ = 2\angle OPQ$ 

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Ans: Given A circle with centre O and an external point T and two tangents TP and TQ to the circle, where P, Q are the points of contact.

To Prove: $\angle PTQ = 2\angle OPQ$ 

Proof: Let $\angle PTQ = \theta $ Since TP, TQ are tangents drawn from point T to the circle.

TP = TQ

$\therefore $ TPQ is an isosceles triangle

$ \therefore \angle TPQ = \angle TQP = \dfrac{1}{2}({{180}^\circ}-\theta )$

$ ={{90}^\circ}-\dfrac{\theta }{2}$

Since, TP is a tangent to the circle at point of contact P

$ \therefore \angle OPT = {{90}^\circ} $

$ \therefore \angle OPQ = \angle OPT - \angle TPQ $

$ = {{90}^\circ} - ( 90 - \dfrac{1}{2}\theta)$

$ = \dfrac{\theta }{2} = \dfrac{1}{2} \angle PTQ$ 

Thus $\angle PTQ = 2 \angle OPQ$ 

13. Prove that the lengths of two tangents drawn from an external point to a circle are equal.

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Ans: Given: P is an external point to the circle C,

PQ and PR are two tangents from P to the circle.

To Prove: PQ = PR

Construction: Join OP

Proof:

$\because $ A tangent to a circle is perpendicular to the radius through the point of contact.

$\therefore \angle OAP={{90}^\circ}=\angle ORP$ 

Now in right triangles POQ and POR,

$OQ = OR$ (Each radius r)

Hypotenuse, OP = Hypotenuse, OP (common)

$ \therefore \vartriangle POQ \cong \vartriangle POR$ (by RHS rule) 

$ \therefore {PQ = PR}$ 

14. The circle of $\vartriangle ABC$ touches the sides BC, CA and AB at D, E and F respectively. If AB= AC, prove that BD = CD.

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Ans: Tangents drawn from an external point to a circle are equal in length

AF = AE (Tangents from A) ...(i)

BF = BD (Tangents from B) ...(ii)

CD = CE (Tangents from C) ...(iii)

Adding (i), (ii)and (iii), we get

AF + BF + CD = AE + BD + CE 

$\Rightarrow$ AB + CD = AC + BD 

But AB = AC (given)

CD = BD 

15. Prove that the tangents drawn at the ends of a chord of a circle make equal angles with chord.

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Ans: Let NM be chord of circle with centre C.

Let tangents at M.N meet at the point O.

Since OM is a tangent

$\therefore OM \bot CM, \angle{OMC} = {{90}^\circ}$ 

$\because ON$ is a tangent

$\therefore ON \bot CN, \angle{ONC} = {{90}^\circ}$ 

Again in $\vartriangle CMN, CM = CN = r$ 

$ \therefore \angle CMN = \angle CNM $

$ \therefore \angle OMC - \angle CMN = \angle ONC - \angle CNM$

$ \Rightarrow \angle OML = \angle ONL $

Thus, tangents make equal angle with the chord.

16. In the given figure, if AB = AC, prove that BE = EC.

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Ans: Since tangents from an exterior point A to a circle are equal in length

$\therefore AD=AF.....(i)$ 

Similarly, tangents from an exterior point B to a circle are equal in length

$\therefore BD=BE......(2)$ 

Similarly, for C

$CF = CF.....(3)$ 

Now $AB = AC$

$\therefore AB - AD = C - AD$ 

$\Rightarrow AB - AD = AC - AF$ .....[by (i)]

$\Rightarrow BD = CF$ 

$\Rightarrow BE = CF$ ......[by (ii)] 

$\Rightarrow$ BE = CE$ [$\because$ BD = BE, CE = CF] [by (iii)] 

17. Find the locus of centre of circle with two intersecting line.

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Ans: Let ${{I}_{1}}, {{I}_{2}}$ be two intersection lines.

Let a circle with centre P touch the two lines ${{I}_{1}},{{I}_{2}}$ and at M and N respectively.

PM = PN (Radii of same circle)

P is equidistant from the lines ${{I}_{1}},{{I}_{2}}$ and

Similarly, centre of any other circle which touch the two intersecting lines ${{I}_{1}},{{I}_{2}}$ will be equidistant from ${{I}_{1}},{{I}_{2}}$.

P lies on a bisector ${{I}_{1}}$ of the angle between ${{I}_{1}},{{I}_{2}}$. 

($\text{The locus of points equidistant from two intersecting lines is the pair of bisectors of the angle between the lines}$)

Hence, locus of centre of circles which touch two intersecting lines is the pair of bisectors of the angles between the two lines.

18. In the given figure, a circle is inscribed in a quadrilateral ABCD. 

If AD = 23 cm, AB = 29 cm and DS = 5 cm, find the radius of the circle.

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Ans: In the given figure, and $OQ$ $\bot$ $BA$ 

Also, $OP=OQ=r$ 

$\therefore OBPQ$ is a square

$\therefore BP = BQ = r$ 

DR = DS = 5cm .....(i)

$ \therefore AR = AD - DR$

= 23 - 5 = 18cm 

AQ = AR = 18cm 

BQ = AB - AQ 

= 20 - 18 = 11cm 

r = 11cm

4 Marks Questions

1. In figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B. Prove that $\angle AOB={{90}^\circ}$.

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Ans: Given: In figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B.

To Prove: $\angle AOB={{90}^{\circ}}$ 

Construction: Join OC

Proof: 

$ \angle OPA = {{90}^\circ}......(i)$ 

$ \angle OCA={{90}^{\circ}}.......(ii)$

 Tangent at any point of a circle is perpendicular to the radius through the point of contact

In right angled triangles OPA and OCA,

OA = OA (Common)

AP = AC ($\text{Tangents from an external point to a circle are equal}$)

$\therefore \vartriangle OPA \simeq \vartriangle OCA$ (RHS congruence criterion)

$ \therefore \angle OAP = \angle OAC$ (By CPCT)

$ \Rightarrow \angle \text{OAC}=\dfrac{1}{2}\angle PAB......(iii)$

Similarly, 

$ \angle OBQ=\angle OBC$

$ \Rightarrow \angle OBC = \dfrac{1}{2}\angle QBA......(iv)$ 

$\because XY \parallel X'Y'$ and a transversal AB intersects them.

$\therefore \angle PAB + \angle QBA = {{180}^\circ}$ ($\text{Sum of the consecutive interior angles on the same side of the transversal is} {{180}^{\circ}}$)

$ \Rightarrow \dfrac{1}{2} + \angle PAB + \dfrac{1}{2}\angle QBA = \dfrac{1}{2}\times {{180}^\circ}......(iv)$ 

$ \Rightarrow \angle OAC + \angle OBC = {{90}^\circ}$

(From eq. (iii) & (iv))

In $\vartriangle AOB$,

$\angle OAC + \angle OBC + \angle AOB = {{180}^\circ}$ 

($\text{Angle sum property of a triangle}$)

$\Rightarrow {{90}^\circ} + \angle AOB = {{180}^\circ}$ 

(From eq. (v))

$\Rightarrow \angle AOB={{90}^\circ}$ 

2. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8cm and 6 cm respectively (see figure). Find the sides AB and AC.

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Ans: Join OE and OF. Also join OA, OB and OC.

Since $BD=8cm$ 

($\text{Tangents from an external point to a circle are equal}$)

Since $CD=6cm$ 

$CF = 6cm$ 

($\text{Tangents from an external point to a circle are equal}$)

$AE = AF = x$ 

Since OD = OE = OF = 4cm 

($\text{Radii of a circle are equal}$)

$\therefore$ Semi-perimeter of $\vartriangle$ ABC

$\dfrac{(x+6)+(x+8)+(6+8)}{2}$ 

$= (x+14)cm$ 

$ \therefore \text{area of }\vartriangle \text{ABC} = \sqrt{s(s-a)(s-b)(s-c)}$ 

$ =\sqrt{(x+14)(x+14-14)(x+14-x+8)(x+14-x+6}$

$ =\sqrt{(x+14)(x)(6)(8)}$

Now, Area of $\vartriangle ABC = area\text{ of }\vartriangle OBC + area\text{ of }\vartriangle OCA + \text{area of} \vartriangle OAB$

$ = \sqrt{(x+14)(x)(6)(8)} {{cm}^{2}}$

$ =\dfrac{(6+8)4}{2}+\dfrac{(x+6)4}{2}+\dfrac{(x+8)4}{2} $

$ \Rightarrow \sqrt{(x+14)(x)(6)(8)}$

$ = 28+2x+12+2x+16 $

$ \Rightarrow \sqrt{(x+14)(x)(6)(8)} $ 

$ = 4x+56 $

$ \Rightarrow \sqrt{(x+14)(x)(6)(8)}=4(x+14) $

Squaring both sides,

$ (x+14)(x)(6)(8)=16{{(x+14)}^{2}} $

$ \Rightarrow 3x=x+14 $

$ \Rightarrow 2x=14 $

$ \Rightarrow x=7 $

$\therefore AB=x+8=7+8=15cm$ 

$\text{and AC=x+6=7+6=13cm}$ 

3. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

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Ans: Given:$ABCD$ is a quadrilateral circumscribing a circle whose centre is O.

To prove: (i) $\angle AOB+\angle COD={{180}^\circ}$ (ii) $\angle BOC+\angle AOD={{180}^\circ}$ 

Construction: Join OP, OQ, OR and OS.

Proof: Since tangents from an external point to a circle are equal.

$ \therefore AP=AS$

BP = BQ ......(i)

CQ = CR 

DR = DS 

In $ \vartriangle OBP\text{ and }\vartriangle \text{OBQ}$

$OP=OQ [\text{Radii of the same circle}]$

$OB=OB$ [Common]

$BP=BQ$ [From eq. (i)]

$\therefore \vartriangle OPB\cong \vartriangle OBQ [\text{By SSS congruence criterion}$]

$\therefore \angle 1=\angle 2$ [By C.P.C.T.]

Similarly, $\angle 3=\angle 4,\angle 5=\angle 6,\angle 7=\angle 8$ 

Since, the sum of all the angles round a point is equal to ${{360}^\circ}$ 

$ \therefore \angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7+\angle 8={{360}^\circ} $

$ \Rightarrow \angle 1+\angle 1+\angle 4+\angle 4+\angle 5+\angle 5+\angle 8+\angle 8={{360}^\circ} $

$ \Rightarrow 2(\angle 1+\angle 4+\angle 5+\angle 8)={{360}^\circ}$

$ \Rightarrow \angle 1+\angle 4+\angle 5+\angle 8={{180}^\circ} $

$ \Rightarrow (\angle 1+\angle 5)+(\angle 4+\angle 8)={{180}^\circ}$ 

$ \Rightarrow \angle AOB+\angle COD={{180}^\circ}$

Similarly, we can prove that

$\angle BOC+\angle AOD={{180}^\circ}$ 

4. In the given figure XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B. Prove that $\vartriangle OAP$

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Ans: Join$OC$ 

In $\vartriangle OAP$ and DAOC, we have

 $AP=AC$ ($\text{tangents from A to the circle are equal}$)

$AO=AO$ 

$OP=OC$ (radius)

$\therefore \vartriangle OAP\cong \vartriangle AOC$ 

$ \therefore \angle 1=\angle 2 $

$ \therefore \angle PAC=2\angle 2 $

Similarly, $\angle CBQ=2\angle 4$ 

Now, $\angle PAC+\angle CBQ={{180}^\circ}[\therefore XY\parallel X'Y']$ 

$ 2\angle 2+2\angle 4={{180}^\circ}$

$ \Rightarrow \angle 2+\angle 4={{90}^\circ} $

But in $\vartriangle AOB$ ,

$ \Rightarrow \angle AOB+\angle OAB+\angle ABO={{180}^\circ}$

$ \Rightarrow \angle AOB+\angle 2+\angle 4={{180}^\circ} $

$ \Rightarrow \angle AOB+{{90}^\circ} $

$ \Rightarrow \angle AOB={{90}^\circ} $

5. In the given figure, $PT$ is tangent and $\angle PAB$ is a secant. If $PT=6cm$ , $AB=5cm$ . Find the length $PA$ 

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Ans: Join $OT$, $OA$, $OP$. Draw $OM$ $\bot $ $AB$ 

Let radius of the circle = r

$\because OT\bot PT[\because $ OT is radius and PT is a tangent]

$\therefore O{{P}^{2}}=P{{T}^{2}}+O{{T}^{2}}$ (From right $\vartriangle$ OPT)

$\Rightarrow O{{P}^{2}}={{6}^{2}}+{{r}^{2}} $

$ \Rightarrow O{{P}^{2}}-{{r}^{2}}=36 $

$ \Rightarrow OP{}^{2}-OA=36....(i)[OA=OT=r] $

Also from right $\vartriangle OMA,$ 

$ O{{A}^{2}}=O{{M}^{2}}+A{{M}^{2}} $

$ \Rightarrow O{{P}^{2}}-36=O{{M}^{2}}+A{{M}^{2}} $

$ \Rightarrow O{{P}^{2}}-O{{M}^{2}}-A{{M}^{2}}=36 $

$ \Rightarrow P{{M}^{2}}-A{{M}^{2}}=36 $

$ \Rightarrow (PM+AM)(PM-AM)=36 $

$ \Rightarrow (PM+AM)PA=36 $

$ \Rightarrow (PM+MB)PA=36 $ 

$[\because AM=MB,\because OM$ bisects AB]

$ (PB)(AP)=36$

$ \Rightarrow PA(PA+AB)=36 $

$ \Rightarrow P{{A}^{2}}+5PA-36=0 $

$ \Rightarrow (PA+9)(PA-4)=0 $

$\Rightarrow PA=4\text{ or }PA=-9$ ($text{It cannot be -ve}$)

6. From a point $P$ two tangents are drawn to a circle with centre $O$ . If $OP$= diameter of the circle, show that $\Delta APQ$ is equilateral.

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Ans: Join $OP$.

Suppose $OP$ meets the circle at $Q$ . Join $AQ$ 

We have

$OP$ = diameter

$OQ+PQ$ = diameter

$PQ$ = Diameter – radius ($\because$ OQ = r)

$PQ=$ radius

Thus, $OQ=PQ=$ radius

Thus, $OP$ is the hypotenuse of right triangle

$OAP$ and $Q$ is the mid-point of $OP$ 

OA = AQ = OQ 

($\text{Mid-point of hypotenuse of a right triangle is equidistant from the vertices}$)

 $\Rightarrow \vartriangle OAQ$ is equilateral

$\vartriangle APB$ $\Rightarrow \angle AOQ={{60}^\circ}$ 

So,

 $ \angle APO={{30}^\circ} $

$ \therefore \angle APB=2\angle APO={{60}^\circ} $ 

Also PA = PB

$ \Rightarrow \angle PAB=\angle PBA $

But

$ \angle APB={{60}^\circ} $

$ \therefore \angle PAB=\angle PBA={{60}^\circ} $

Hence $\Delta APB$ is equilateral.

CBSE Important Questions Class 10 Maths Chapter Wise with Solutions PDF

You simply need to click on the name of the chapter for which you are looking to revise the important questions for Class 10 Maths from the PDF link available at the official website of Vedantu. You will be directed to the page for the particular chapter you want which includes all the important questions along with solutions. Spare time out to practice for chapter 10 maths class 10 important questions provided by Vedantu, if you really seek to score excellent marks.

Class 10 Maths Chapter Wise Questions With Solutions

Solutions available are a detailed, step-by-step depiction that makes it easier for you to understand the problems and how to solve them. The important formulas and concepts for each chapter are also provided. These will be extremely helpful while revising the chapter.

CBSE Important Questions With Solutions For Class 10 Maths Chapter Wise

Here, you will find the list of the NCERT important questions for Class 12 Maths PDF for every chapter in the CBSE syllabus of Maths. A step-by-step solution for each question has also been provided. Expert maestros of Mathematics have solved these Class 12 Maths Important Questions chapters-wise with answer keys for you to make your preparation easier.

Class 10 Maths Circles Important Questions With Solutions

Questions available at Vedantu in the section of Class 10 Maths Ch 10 Important Questions may not be asked directly in the CBSE Exam 2024. However, board students may expect similar kinds of questions since it includes important questions that are repeatedly asked. To solve all questions correctly, you would require to be very well acquainted with the important topics and concepts of the chapter. For this, work on these free Chapter 10 Maths Class 10 Important Questions and other practice test questions.

Benefits Of Class 10 Maths Circles Important Questions With Solutions

Chapter 10 Maths Class 10 Important Questions of Circles provided at the Vedantu based on the new pattern of CBSE. Students preparing for Class 10 Board exams can practice these important questions of Circles in order to gain full marks for the questions from this chapter. Students may meet up with a few of these questions in the 10th board exam. So, they should practice them well and provide 100% for the exam preparation.

Class 10 Maths Circles Important Questions

Chapter 10 Circles of CBSE class 10 has less number of questions. There is a field of reference that objective-based questions, including long answer and short answer questions, will appear in the board examination. Students are recommended to refer to these Important Questions for Class 10 Maths as an element of their board exam preparation. In addition to the important questions, a detailed explanation is also given. Students can refer to it anytime, anywhere when they find themselves stuck with the question.

Important Resources For Preparation of Class 10 CBSE Board Exam

Class 10 students can find revision notes and NCERT solutions prepared by Vedantu experts that will help them in preparation of their CBSE Board Exam. Students can check the below-given links to find the necessary articles in order to prepare in the most effective manner and perform excellently in the exam:

CBSE Class 10 Maths Circles Important Questions With Solutions

• CBSE Class 10 Exam Pattern 2024: Mathematics

• CBSE Class 10 Syllabus 2024: Mathematics

• CBSE Class 10 Board Exam Sample Papers with Marking Scheme 2024: Mathematics

• CBSE Class 10 Question Papers 2019 with Solutions: Mathematics

• CBSE Class 10 Question Papers 2019 with Answer Keys: Mathematics

• CBSE Class 10 NCERT Exemplar Problems, New Textbooks & Solutions for Maths

Solved Practice Problems & Important Questions For Class 10 Maths Circles

Below are a few examples of class 10 maths ch 10 important questions with step-by-step solutions and answer keys.

Example:

In the figure shown below, O is the centre point of a circle, PQ is a chord in the circle and PT is the tangent at P. If ∠POQ measures 70°, then calculate angle ∠TP.

(Image to be added soon)

Solution:

Given:

∠POQ = 70°

From the triangle PQO: ∠1 = ∠2

∠1 + ∠2 + 70° = 180°

∠1 + ∠2 = 180° – 70°

2∠1 = 110°

⇒ ∠1 = 55°

∠1 + ∠TPQ = 90°

Substituting the values, we get:

55° + ∠TPQ = 90°

⇒ ∠TPQ = 90° – 55°

 = 35°

(Image to be added soon)

Example:

In the figure shown below, AB and AC are tangents with centre o to the circle in a way that ∠BAC = 40°. Then, find out ∠BOC.

(Image to be added soon)

Solution:

Given: AB and AC are tangents to the circle

Therefore, ∠AOB = ∠AOC = 90°

In ABOC,

∠AOB + ∠AOC + ∠ABC + ∠BOC = 360°

90° + 90° + 40° + ∠BOC = 360°

∠BOC = 360 – 220°

= 140°

(Image to be added soon)

Practice Questions of Class 10 Maths - Chapter 10 Circles

Following are some of the questions that students can practise to score good marks in the upcoming boards examinations.

Question 1

A 10 cm radius circle's chord forms a straight angle at its centre. Calculate the chord's length in cm.

Answer: AB = 102 cm is the correct answer.

Question 2

Demonstrate that the tangents drawn at the extremities of a circle's diameter are parallel.

Question 3

BC = 12 cm and AB = 5 cm in a right triangle ABC, right-angled at B. Calculate the circumference of the circle encircling the triangle.

Answer: 13 centimetres is the answer.

Question 4

Find the length of OP if the angle between two tangents drawn from an exterior point P to a circle of radius a and centre O is 60°.

Related Study Materials

Following are some of the useful tools that can be used by the students to excel in the board examinations. 

• NCERT Solutions for Class 10 Maths Chapter 10 Circles

• Revision Notes for Circles of Chapter 10

Class 10 Maths Ch 10 Short Answer Type Important Questions

Q1: If a right-angled triangle ABC is right-angled at B. AB = 5 cm and BC = 12 cm. Then find out the radius of the circle that is inscribed in triangle ABC?

Answer: The radius of the circle is 2 cm.

Q2: Given: A triangle XYZ that is an isosceles triangle and YZ is tangent to the circle with centre X. Find out what will be ∠XYZ?

Answer: The measure of ∠XYZ in the given isosceles triangle XYZ will be 45°.

Q3: What will be the angle between the two tangents to the circle which is constructed at the end of two radii and are disposed of at an angle measurement of 45°?

Answer: The angle between them should be 135°.

Class 10 Maths Ch 10 Long Answer Type Important Questions

Following are the type of long type circles class 10 important questions that you might come across while learning and attempting the class 10maths board exam. 

Q.1: From point B, the length of the tangent to a circle measures 24 cm and the distance of B from the centre is 25 cm. What will be the radius of the circle?

(A) 15 cm (B) 25.5 cm

(C) 7cm (D) 12 cm

Solution: First, we would need to draw a perpendicular from the centre O of the triangle to a point B on the circle which touches the tangent. See that this line will be perpendicular to the tangent.

(Image to be added soon)

Thus, OB is perpendicular to AB i.e. OB ⊥ AB

it can also be observed that △OAB is a right-angled triangle.

And we are given that:

AB = 24 cm

OB = 25 cm

Now, applying the Pythagorean theorem in △OAB,

We get

OB2 = OA2 + AB2

=> (25)2 = OB2 + (24)2

= 625 = OB2 + 576

=> OB2 = 625 – 576

=> OB2 = 49

=> OB= 7 cm

Thus, option C i.e. 7 cm is the radius of the given circle.

Exemplary Practice Questions For Class 10 Maths Circles Important Questions

Q. Prove that the parallelogram encompassing a circle is a rhombus.

Q. In a right triangle PQR in which ∠P = 90°, a circle is constructed with PQ as diameter bisecting the hypotenuse PR and P. Prove that the tangent to the circle at O intersects BC.

Q. How many tangents can a circle contain?

Q. If an isosceles triangle MNO, in which MN = NO = 6 cm, is circumscribed in a circle of radius measuring 9 cm. calculate the area of the triangle.

Significance of Class 10 Maths Circles Important Questions

A circle is a closed geometric figure with a lot of geometric elements in it to study. Each of these elements has specific features that need to be studied properly. There are mathematical methods to determine the features of these elements such as the radius, diameter, chord, circumference, sector, centre, etc. This chapter has been included in the syllabus of Class 10 Mathematics to develop the conceptual foundation among the students.

To make this chapter easier to comprehend and to develop problem-solving skills, the experts of Vedantu have compiled a set of important questions based on the latest syllabus. The questions have been set as per the CBSE standards so that students can correlate the concepts well. Solving these questions will automatically escalate the concept level and deepen the knowledge related to this chapter.

This set of questions also comes with respective solutions. These solutions will act as a guide to solving these questions and learning how to use the concepts taught in this chapter. Hence, using these important questions as practice sets will be helpful for the students to prepare this chapter well.

Benefits of Class 10 Maths Circles Important Questions with Solutions

Explore essential questions for Class 10 Maths Chapter 10 (Circles) on Vedantu, tailored to CBSE's new pattern. These crucial questions, designed for Class 10 Board exam preparation, offer an opportunity to secure full marks in this chapter. Given the likelihood of encountering similar questions in the 10th board exam, thorough practice is recommended for optimal exam readiness, aiming for a 100% preparedness level.

Advantages of Class 10 Maths Chapter 10 Circles Important Questions

• The questions have been formulated by the experts by following the trends of the CBSE board in setting exam questions. Hence, these suggested questions will broaden your preparation level for this chapter.

• Solving these questions once the chapter is completely prepared will help you to develop a unique strategy to attempt to answer all questions within the limited exam time.

• You will also be able to define the precise methods of solving questions related to circles and focus on your conceptual development. These methods help you to compile the right answers and score well in the exams.

• Resolve doubts related to these questions faster and focus on how to become better at solving questions for this chapter.

Download CBSE Class 10 Maths Circles Important Questions and Answers PDF

Get the free PDF version of this set of questions and answers and practice. Complete preparing this chapter and understand all the concepts and geometric principles explained first and then use these important questions as an assessment tool. Find out which part of this chapter needs more attention and prepare well.

CBSE Class 10 Maths Circles Important Questions With Solutions

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