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 $\mathbf{AB}\mathbf{+}\mathbf{CD}\mathbf{=}\mathbf{AD}\mathbf{=}\mathbf{BC}$ . 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

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

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

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

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

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

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

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

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

$={\displaystyle \frac{9}{16}}\times 64$

$=36$

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 $\mathrm{\angle }(APB)={60}^{\circ }$. 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 $\mathrm{\angle }APB={60}^{\circ }$. 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) $\sqrt{\mathbf{119}}\mathbf{cm}$ 

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

(The tangent at any point of a circle is $\mathrm{\perp }$ 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 $\mathrm{\angle }POQ={110}^{\circ }$, then $\mathrm{\angle }PTQ=$ 

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1. ${\mathbf{60}}^{\circ }$ 

2. ${\mathbf{70}}^{\circ }$ 

3. ${\mathbf{80}}^{\circ }$ 

4. ${\mathbf{90}}^{\circ }$ 

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

In quadrilateral $OPTQ$,

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

(Angle sum property of quadrilateral)

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

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

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

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 $\mathrm{\angle }POA$ is equal to:

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

(B) ${\mathbf{60}}^{\circ }$ 

(C) ${\mathbf{70}}^{\circ }$ 

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

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

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

In $△OPA$ ,

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

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

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

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.

$\therefore \mathrm{\angle }OPA={90}^{\circ }$…(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.

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

From eq. (i) and (ii),

$\mathrm{\angle }OPA=\mathrm{\angle }OQD$ 

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.

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

$\therefore O{A}^2=O{P}^2+A{P}^2$ (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, $\mathrm{\angle }OPA={90}^{\circ }$ 

(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 $OP\mathrm{\perp }AB$ 

OP bisects AB

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

$△OAP,O{A}^2=O{P}^2+A{P}^2$

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

$\Rightarrow OP=4cm$

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 $△ADB$ and $△ADC$ 

BD = DC

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

AD = AD (common) 

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

$\therefore \mathrm{\angle }ABD=\mathrm{\angle }ACD$ (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 $O^{\prime }$ which touches the line $AB$ at $P$ .

Thus,

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

$\mathrm{\angle }{O}^{\prime }PB={90}^{\circ }$ 

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,

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

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

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

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

As all the angles of the triangle PAB are equal,

$\therefore \mathrm{\Delta }PAB$ 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 $△APQ$ 

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

= 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 $\mathrm{\Delta }ABC$ 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 $\mathrm{\Delta }ABC=AB+BC+AC$ 

= (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.

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

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

$\Rightarrow 25+5x=64$

$\Rightarrow x={\displaystyle \frac{39}{8}}=7.8cm$

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 ${\displaystyle \frac{CA}{OD}}$.

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Ans: Since $BC$ is a diameter

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

also OD $\mathrm{\perp }$ AB

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

Considering the triangles $\mathrm{\Delta }ACB,\mathrm{\Delta }DOB$ we have,

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

$\mathrm{\angle }ABC=\mathrm{\angle }DBC$

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

$\mathrm{\angle }ACB=\mathrm{\angle }DOB$

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

So, the sides will be in proportion that is,

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

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,

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

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 $\mathrm{\angle }OCQ={40}^{\circ }$ and $\mathrm{\angle }ABD={60}^{\circ }$ ,Find $\mathrm{\angle }BCP$ 

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Ans: BD is a diameter

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

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

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

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

In right $△OTA={90}^{\circ }$, 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: $\mathrm{\angle }OPA={90}^{\circ }....(i)$

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

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

OAPB is quadrilateral.

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

(Angle sum property of a quadrilateral)

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

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

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

$\therefore \mathrm{\angle }APB$ and $\mathrm{\angle }AOB$ 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 $\mathrm{\angle }AOB$?

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

$\therefore OA\mathrm{\perp }PA,OB\mathrm{\perp }PB$

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

$\therefore$ Quadrilateral PAOB is cyclic 

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

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

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

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.

$\therefore OM\mathrm{\perp }PQ$ 

$\because$ PQ is a chord of the bigger circle and OM $\mathrm{\perp }$ PQ 

$\therefore OM\text{ bisects PQ}$

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

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

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

 $\text{=}\sqrt{25-9}=4$

$\therefore$ 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]

$\therefore OT\mathrm{\perp }PQ$ 

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

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

$△TRP\sim △TRQ$ (by AA similarity)

$\therefore {\displaystyle \frac{TP}{PO}}={\displaystyle \frac{RP}{RO}}$

$\Rightarrow {\displaystyle \frac{TP}{5}}={\displaystyle \frac{4}{3}}$

$\Rightarrow TP={\displaystyle \frac{20}{3}}cm$

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 = ${\displaystyle \frac{1}{2}}$ (perimeter of $△ABC$).

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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 $△$ ABC = AB + BC + AC 

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

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

= AQ = AR = ${\displaystyle \frac{1}{2}}$ perimeter of $△$ 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

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

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

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

In ACP and BCP we have

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

PC = PC (Common)

$\mathrm{\angle }APO=\mathrm{\angle }BPO$ (proved)

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

$\therefore AC=CB$ (CPCT)

And $\mathrm{\angle }ACP=\mathrm{\angle }BCP$ (CPCT)

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

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

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 $\mathrm{\angle }TRQ={30}^{\circ }$ , find $\mathrm{\angle }PRS$ 

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

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

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

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

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

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

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.

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

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

$\therefore (AP+DP)+(BR+CR)$ 

$=AQ+DS+BQ+CS$ 

$=(AQ+BQ)+(CS+DS)$ 

$\Rightarrow AD+BC=AB+BC$ 

$\Rightarrow BC+BC=AB+AB$

$[\because AB=DC,AD=BC]$ 

$\Rightarrow 2BC=2AB$ 

$\Rightarrow BC=AB$ 

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:

$(i)\mathrm{\angle }AOP=\mathrm{\angle }BOQ$

$(ii)\mathrm{\angle }OPA=\mathrm{\angle }OPB$

Construction: Join AO, PO and BO

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

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

AO = BO (Radii of the same circle)

PO = PO (common)

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

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

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

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

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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: $\mathrm{\angle }PTQ=2\mathrm{\angle }OPQ$ 

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

TP = TQ

$\therefore$ TPQ is an isosceles triangle

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

$={90}^{\circ }-{\displaystyle \frac{\theta }{2}}$

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

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

$\therefore \mathrm{\angle }OPQ=\mathrm{\angle }OPT-\mathrm{\angle }TPQ$

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

$={\displaystyle \frac{\theta }{2}}={\displaystyle \frac{1}{2}}\mathrm{\angle }PTQ$ 

Thus $\mathrm{\angle }PTQ=2\mathrm{\angle }OPQ$ 

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.

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

Now in right triangles POQ and POR,

$OQ=OR$ (Each radius r)

Hypotenuse, OP = Hypotenuse, OP (common)

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

$\therefore PQ=PR$ 

14. The circle of $△ABC$ 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

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

$\because ON$ is a tangent

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

Again in $△CMN,CM=CN=r$ 

$\therefore \mathrm{\angle }CMN=\mathrm{\angle }CNM$

$\therefore \mathrm{\angle }OMC-\mathrm{\angle }CMN=\mathrm{\angle }ONC-\mathrm{\angle }CNM$

$\Rightarrow \mathrm{\angle }OML=\mathrm{\angle }ONL$

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 $OQ$ $\mathrm{\perp }$ $BA$ 

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 $\mathrm{\angle }AOB={90}^{\circ }$.

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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: $\mathrm{\angle }AOB={90}^{\circ }$ 

Construction: Join OC

Proof: 

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

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

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

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

$\therefore \mathrm{\angle }OAP=\mathrm{\angle }OAC$ (By CPCT)

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

Similarly, 

$\mathrm{\angle }OBQ=\mathrm{\angle }OBC$

$\Rightarrow \mathrm{\angle }OBC={\displaystyle \frac{1}{2}}\mathrm{\angle }QBA......(iv)$ 

$\because XY\parallel X^{\prime }{Y}^{\prime }$ and a transversal AB intersects them.

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

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

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

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

In $△AOB$,

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

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

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

(From eq. (v))

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

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 $△$ ABC

${\displaystyle \frac{(x+6)+(x+8)+(6+8)}{2}}$ 

$=(x+14)cm$ 

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

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