NCERT Solutions for Class 10 Maths Chapter 10 Circles

The NCERT Solutions for Class 10 Maths Chapter 10 Circles comprise two exercises, Ex-10.1 and Ex-10.2. Solving the sums given in these two exercises is going to provide students with a deeper understanding of the concepts of circles, various theorems, corollaries, and applications of the formulas of Circles.

Exercise 10.1: The first exercise of Class 10 Maths Chapter 10 Circles is a short one, consisting of only 4 sums. These sums will help students to assess their understanding of the properties of circles, including the concepts of tangents, secants, radii, diameters, chords, etc. There are fill-in-the-blank and short answer-type questions covered in Ex-10.1. Students will have to apply the Pythagoras theorem in solving one of the sums of this exercise, while the last sum is based on the construction of circles.

Exercise 10.2: The first exercise consists of a total of 13 sums. Students will be able to get a good understanding of various concepts of angles, triangles, quadrilaterals, circles inscribed in quadrilaterals, circles circumscribing polygons, Pythagoras theorem, etc., when they solve the sums given in this exercise. Students will need to calculate the lengths and angles subtended by chords, tangents, etc., for some of the sums. The remaining sums of this exercise are mostly based on the deduction of equations using the application of various formulas and theorems of circles. The solutions to the sums given in NCERT Class 10 Maths Exercise 10.2 are very important from the exam point of view.

Access NCERT Solutions for Class 10 Maths Chapter 10 – Circles

Exercise 10.1

1. How many tangents can a circle have?

Ans:

We know that a circle has an infinite number of points on its perimeter. Hence, there will be an infinite number of tangents on a circle.

2. Fill in the blanks: 

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

Ans:

A tangent to a circle intersects it in exactly one point(s).

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

Ans:

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

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

Ans:

A circle can have two parallel tangents at the most.

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

Ans:

The common point of a tangent to a circle and the circle is called point of contact.

3. A tangent $$PQ$$ at a point $$P$$ of a circle of radius $$5\text{cm}$$ meets a line through the centre $$O$$ at a point $$Q$$ so that $$OQ=12\text{cm}$$. 

Length of $$PQ$$ is: (A) $$12\text{cm}$$ (B) $$13\text{cm}$$ (C) $$8.5\text{cm}$$ (D) $$\sqrt{119}$$.

Ans:

As, the tangent $PQ$ is base and radius is the height of $5\text{cm}$. Also, $OQ$ is the hypotenuse. Therefore, the length of $PQ$ will be –

$PQ=\sqrt{O{Q}^2-O{P}^2}$

$$\Rightarrow PQ=\sqrt{{12}^2-5^2}$$

$$\Rightarrow PQ=\sqrt{144-25}$$

$$\Rightarrow PQ=\sqrt{119}$$

Hence, option $$(D)$$ is correct.

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

Ans:

From the figure given above let us assume that a line $$l$$ and a circle having centre $$O$$, comprises a line $$PT$$ which is parallel to the line $$l$$ and is a tangent to the circle. Similarly, $$AB$$ is a secant parallel to the line $$l$$.

Exercise 10.2

1. From a point $$Q$$, the length of the tangent to a circle is $$24\text{cm}$$ and the distance of $$Q$$ from the centre is $$25\text{cm}$$. The radius of the circle is 

(A) $$7\text{cm}$$ (B) $$12\text{cm}$$ (C) $$15\text{cm}$$ (D) $$24.5\text{cm}$$.

Ans:

From the figure given above, let us assume that $$O$$ is the centre of the circle and given we have $PQ$ as the tangent to the circle of length $24\text{cm}$, and $OQ$ is of length $25\text{cm}$.

Now, by using the Pythagoras theorem we have –

$OP=\sqrt{O{Q}^2-P{Q}^2}$

$\Rightarrow OP=\sqrt{{25}^2-{24}^2}$

$\Rightarrow OP=\sqrt{625-576}$

$\Rightarrow OP=\sqrt{49}$

$\Rightarrow OP=7\text{cm}$.

Hence, option $$(A)$$ is correct.

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

(A) $$60{}^{\circ }$$ (B) $$70{}^{\circ }$$ (C) $$80{}^{\circ }$$ (D) $$90{}^{\circ }$$.

Ans:

As, from the given figure, we can observe that there are two tangents perpendicular to the radius of the circle as $TP$ and $TQ$. Since, they are perpendicular to the radius hence, we have $\mathrm{\angle }OPT=90{}^{\circ }$ and 

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

Therefore, $\mathrm{\angle }PTQ=360{}^{\circ }-\mathrm{\angle }OPT-\mathrm{\angle }OTQ-\mathrm{\angle }POQ$

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

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

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

Hence, option $$(B)$$ is correct.

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

(A) $$50{}^{\circ }$$ (B) $$60{}^{\circ }$$ (C) $$70{}^{\circ }$$ (D) $$80{}^{\circ }$$.

Ans:

As, we have two tangents $PA$ and $PB$ which will be perpendicular to the radius of the circle. Hence, we have $\mathrm{\angle }PAO=90{}^{\circ }$ and 

$\mathrm{\angle }PBO=90{}^{\circ }$. Since this is a quadrilateral, we will have the sum of all interior angles equal to $360{}^{\circ }$.

Therefore,

$\mathrm{\angle }PAO+\mathrm{\angle }PBO+\mathrm{\angle }APB+\mathrm{\angle }AOB=360{}^{\circ }$

$\Rightarrow 90{}^{\circ }+90{}^{\circ }+80{}^{\circ }+\mathrm{\angle }AOB=360{}^{\circ }$

$\Rightarrow \mathrm{\angle }AOB=360{}^{\circ }-260{}^{\circ }$

$\Rightarrow \mathrm{\angle }AOB=100{}^{\circ }$.

Now, we know that $\mathrm{\angle }POA$ is half of the $\mathrm{\angle }AOB$.

Therefore,

$\mathrm{\angle }POA=\frac{\mathrm{\angle }AOB}{2}$

$\Rightarrow \mathrm{\angle }POA=\frac{100{}^{\circ }}{2}$

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

Hence, option $$(A)$$ is correct.

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

Ans:

As, from the figure given above, let us assume that $$O$$ is the centre of the circle and $$AB$$, $$CD$$ are the two tangents at the ends of the diameter of the circle.

Now, we know that tangents are perpendicular to the radius of the circle.

Hence, $$\mathrm{\angle }CQO=90{}^{\circ }$$

$$\mathrm{\angle }DQO=90{}^{\circ }$$

$$\mathrm{\angle }APO=90{}^{\circ }$$

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

Therefore, we can say that $$\mathrm{\angle }CPQ=\mathrm{\angle }BQP$$ because they are alternate angles. Similarly, $$\mathrm{\angle }AQP=\mathrm{\angle }QPD$$.

Hence, if the interior alternate angles are equal then the lines $$AB$$, $$CD$$ should be parallel lines.

Hence, proved.

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

Ans:

Let us assume that the line perpendicular at the point of contact to the tangent of a circle does not pass through the centre $$O$$ but passes through a point $$Q$$ as shown in the figure above.

Hence, we have the line $$PQ\mathrm{\perp }AB$$.

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

Also, we have $\mathrm{\angle }OPB=90{}^{\circ }$.

After comparing both the equations we get –

$\Rightarrow \mathrm{\angle }QPB=\mathrm{\angle }OPB$

But, from the figure drawn above we can observe that this is not the case as $\mathrm{\angle }QPB<\mathrm{\angle }OPB$.

Therefore, we can conclude that $\mathrm{\angle }QPB\ne \mathrm{\angle }OPB$. They can only be equal when these two-line segments $QP$ and $OP$ will be equal. This implies that the line perpendicular at the point of contact to the tangent of a circle passes through the centre $$O$$. Hence, proved.

6. The length of a tangent from a point $$A$$ at distance $$5\text{cm}$$ from the centre of the circle is $$4\text{cm}$$. Find the radius of the circle.

Ans:

From the figure given above, let us assume that $$O$$ is the centre of the circle and given we have $AB$ as the tangent to the circle of length $4\text{cm}$, and $OA$ is of length $5\text{cm}$.

Now, by using the Pythagoras theorem we have –

$OB=\sqrt{O{A}^2-A{B}^2}$

$\Rightarrow OB=\sqrt{5^2-4^2}$

$\Rightarrow OB=\sqrt{25-16}$

$\Rightarrow OB=\sqrt{9}$

$\Rightarrow OB=3\text{cm}$.

Therefore, the radius of the circle will be $3\text{cm}$.

7. Two concentric circles are of radii $$5\text{cm}$$ and $$3\text{cm}$$. Find the length of the chord of the larger circle which touches the smaller circle.

Ans:

From the figure given above, we can observe that the line segment $PQ$ is the chord of the larger circle and a tangent to the smaller circle.

Therefore, $$OA\mathrm{\perp }PQ$$.

Now, we can observe that the $$\mathrm{\Delta }OAP$$ forms a right-angled triangle.

Hence, by applying Pythagoras theorem –

$$AP=\sqrt{O{P}^2-O{A}^2}$$

$$\Rightarrow AP=\sqrt{5^2-3^2}$$

$$\Rightarrow AP=\sqrt{25-9}$$

$$\Rightarrow AP=\sqrt{16}$$

$$\Rightarrow AP=4\text{cm}$$.

Now, since the radius is perpendicular to the tangent therefore, we have $$AP=AQ$$.

Hence, $$PQ=2AP$$.

$$\Rightarrow PQ=2\times 4$$

$$\Rightarrow PQ=8\text{cm}$$.

So, the length of the chord will be of $$8\text{cm}$$.

8. A quadrilateral $$ABCD$$ is drawn to circumscribe a circle. Prove that $$AB+CD=AD+BC$$.

Ans:

From the figure given we can observe that the sides of the quadrilateral acts as tangents to the circle. As, the tangents drawn from any external point will have the same length. Therefore, we have –

$$DR=DS$$, 

$$CR=CQ$$, 

$$BP=BQ$$, and 

$$AP=AS$$.

Now, we will add all the relations.

Hence, $$DR+CR+BP+AP=DS+CQ+BQ+AS$$

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

$$\Rightarrow DC+AB=AD+BC$$

Hence, proved.

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

Ans:

From the figure given we can observe that $$AB$$, $$XY$$ and $$X^{\prime }{Y}^{\prime }$$ are tangents to the circle and will be perpendicular to its radius.

Now, let us consider two triangles $$\mathrm{\Delta }OPA$$ and $$\mathrm{\Delta }OCA$$, such that –

$$OP=OC$$,

$$AP=AC$$.

Hence, we have $$\mathrm{\Delta }OPA\cong \mathrm{\Delta }OCA$$ by the SSS congruence rule.

$$\Rightarrow \mathrm{\angle }POA=\mathrm{\angle }COA$$.

In the similar manner $$\mathrm{\Delta }OQB\cong \mathrm{\Delta }OCB$$.

$$\Rightarrow \mathrm{\angle }QOB=\mathrm{\angle }COB$$.

Now, we know that $$PQ$$ is the diameter of the circle.

$$\Rightarrow \mathrm{\angle }POA+\mathrm{\angle }COA+\mathrm{\angle }COB+\mathrm{\angle }QOB=180{}^{\circ }$$

$$\Rightarrow 2\mathrm{\angle }COA+2\mathrm{\angle }COB=180{}^{\circ }$$

$$\Rightarrow \mathrm{\angle }COA+\mathrm{\angle }COB=90{}^{\circ }$$

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

Hence, proved.

10. 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.

Ans:

Let us assume a circle with centre at $$O$$, which have two tangents $$PA$$ and $$PB$$ which are perpendicular to the radius of the circle as shown in the figure above.

Now, let us consider two triangles $$\mathrm{\Delta }OAP$$ and $$\mathrm{\Delta }OBP$$, such that –

$$PA=PB$$, and 

$$OA=OB$$.

Hence, we have $$\mathrm{\Delta }OAP\cong \mathrm{\Delta }OBP$$ by the SSS congruence criteria.

Therefore, $$\mathrm{\angle }OPA=\mathrm{\angle }OPB$$ and

$$\mathrm{\angle }AOP=\mathrm{\angle }BOP$$.

This implies that $$\mathrm{\angle }APB=2\mathrm{\angle }OPA$$ and

$$\mathrm{\angle }AOB=2\mathrm{\angle }AOP$$.

Hence, in the right-angled triangle $$\mathrm{\Delta }OAP$$, we have –

$$\mathrm{\angle }AOP+\mathrm{\angle }OPA=90{}^{\circ }$$

$$2\mathrm{\angle }AOP+2\mathrm{\angle }OPA=180{}^{\circ }$$

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

$$\Rightarrow \mathrm{\angle }AOP+\mathrm{\angle }OPA=180{}^{\circ }$$.

Hence, proved.

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

Ans:

From the figure given above, we can observe that $$ABCD$$ is a parallelogram whose sides are tangent to the circle. This implies that $$DR=DS$$, 

$$CR=CQ$$, 

$$BP=BQ$$, and 

$$AP=AS$$.

Now, we will add all the relations.

Hence, $$DR+CR+BP+AP=DS+CQ+BQ+AS$$

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

$$\Rightarrow DC+AB=AD+BC$$

As, the sides of a parallelogram are always parallel and equal in length.

Therefore, $$AB=DC$$ and

$$AD=BC$$.

$$\Rightarrow 2AB=2BC$$

$$\Rightarrow AB=BC$$

Hence, all the sides of parallelogram are equal.

Therefore, we can conclude that it is a rhombus. Hence, proved.

12. A triangle $$ABC$$ is drawn to circumscribe a circle of radius $$4\text{cm}$$ such that the Segments $$BD$$ and $$DC$$ into which $$BC$$ is divided by the point of contact $$D$$ are of lengths $$8\text{cm}$$ and $$6\text{cm}$$ respectively. Find the sides $$AB$$ and $$AC$$.

Ans:

From the figure given above, we can observe that the sides of a triangle $$ABC$$ are the tangents to the circle and are perpendicular to the radius of the circle.

Hence, in $$\mathrm{\Delta }ABC$$

$$CF=CD=6\text{cm}$$

Similarly, $$BE=BD=8\text{cm}$$ and

$$AE=AF=x\text{cm}$$.

In $$\mathrm{\Delta }ABE$$,

$$AB=AE+BE$$

$$\Rightarrow AB=x+8$$

Similarly, $$BC=8+6=14$$ and

$$CA=6+x$$.

Therefore, we have $$2s=AB+BC+CA$$

$$\Rightarrow 2s=x+8+14+6+x$$

$$\Rightarrow 2s=2x+28$$

$$\Rightarrow s=x+14$$.

Now, we know that the area of a triangle can be calculated by using the formula $$Area=\sqrt{s(s-a)(s-b)(s-c)}$$.

Therefore,

$$Area\text{of}\mathrm{\Delta }\text{ABC}=\sqrt{(14+x)(14+x-6-x)(14+x-8-x)}$$

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

$$\Rightarrow 4\sqrt{3(14x+x^2)}$$.

Hence, $$Area\text{of}\mathrm{\Delta }\text{OBC = }\frac{1}{2}\times OD\times BC$$

$$Area\text{of}\mathrm{\Delta }\text{OBC =}28$$

$$\Rightarrow Area\text{of}\mathrm{\Delta }\text{OCA = }\frac{1}{2}\times OF\times AC$$

$$Area\text{of}\mathrm{\Delta }\text{OCA = 12+2x}$$

$$\Rightarrow Area\text{of}\mathrm{\Delta }\text{OAB = }\frac{1}{2}\times OE\times AB$$

$$Area\text{of}\mathrm{\Delta }\text{OAB = 16+2x}$$

Therefore, the total area of the triangle $$ABC$$ will be –

$$=Area\mathrm{\Delta }\text{OBC}+Area\mathrm{\Delta }\text{OCA+}Area\mathrm{\Delta }\text{OAB}$$

$$\Rightarrow 4\sqrt{3(14x+x^2)}=28+12+2x+16x+2x$$

$$\Rightarrow \sqrt{3(14x+x^2)}=14+x$$

$$\Rightarrow 3(14x+x^2)={(14+x)}^2$$

Hence, after further solving we have –

$$(x+14)(x-7)=0$$

$$\Rightarrow x=-14$$ or

$$\Rightarrow x=7$$

As, the side cannot be negative in nature, therefore, $$x=7\text{cm}$$.

Hence, $$AB=7+8$$

$$\Rightarrow AB=15\text{cm}$$ and

$$CA=6+7$$

$$\Rightarrow CA=13\text{cm}$$.

13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend Supplementary angles at the centre of the circle.

Ans:

From the figure given we can observe that the sides of the quadrilateral acts as tangents to the circle. Therefore, we have –

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

$$\mathrm{\angle }BOC+\mathrm{\angle }DOA=180{}^{\circ }$$

Now, let us consider two triangles $$\mathrm{\Delta }OAP$$ and $$\mathrm{\Delta }OAS$$,

Hence, we have $$AP=AS$$, and

$$OP=OS$$

$$\Rightarrow \mathrm{\Delta }OAP\cong \mathrm{\Delta }OAS$$ by the SSS congruence criteria.

Thus, $$\mathrm{\angle }POA=\mathrm{\angle }AOS$$

$$\mathrm{\angle }1=\mathrm{\angle }8$$

Also,

$$\mathrm{\angle }2=\mathrm{\angle }3$$,

$$\mathrm{\angle }4=\mathrm{\angle }5$$,

$$\mathrm{\angle }6=\mathrm{\angle }7$$

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

$$\Rightarrow (\mathrm{\angle }1+\mathrm{\angle }8)+(\mathrm{\angle }2+\mathrm{\angle }3)+(\mathrm{\angle }4+\mathrm{\angle }5)+(\mathrm{\angle }6+\mathrm{\angle }7)=360{}^{\circ }$$

$$\Rightarrow 2\mathrm{\angle }1+2\mathrm{\angle }2+2\mathrm{\angle }5+2\mathrm{\angle }6=360{}^{\circ }$$

$$\Rightarrow (\mathrm{\angle }1+\mathrm{\angle }2)+(\mathrm{\angle }5+\mathrm{\angle }6)=180{}^{\circ }$$

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

In the similar manner we can prove that $$\mathrm{\angle }BOC+\mathrm{\angle }DOA=180{}^{\circ }$$.

Therefore, we have proved that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

NCERT Solutions for Class 10 Maths Chapter 10 Exercises

Conclusion

The NCERT Solutions for Class 10 Maths Chapter 10 - Circles, provided by Vedantu, are designed to help students understand the properties and theorems related to circles. It's important to focus on key concepts like tangent properties, the number of tangents from a point to a circle, and the theorems involving angles and radii. These concepts are crucial for solving problems accurately. From previous year question papers, around 2 to 3 questions are typically asked from this chapter. Understanding and practicing these solutions will help students score well in their exams.

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NCERT Solutions for Class 10 Maths - Other Chapters

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Study Resources for Class 10 Maths

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