NCERT Solutions Class 10 Maths Chapter 10 – Circles Exercise 10.2

NCERT Solutions for Class 10 Maths Chapter 10 – Circles Exercise 10.2 are designed to help students understand and apply the properties of tangents, secants, and chords in a circle. This exercise involves solving problems related to angles subtended by chords at the center, properties of cyclic quadrilaterals, and intersecting chords.

Exercise 10.2 covers:

NCERT Solutions Class 10 Maths Chapter 10 – Circles Exercise 10.2

Circles

Medium

Q.

$\begin{array}{l}\text{In the following figure, if TP and TQ are the two tangents to a circle with centre O so that }\\ \angle \text{POQ}=\text{110°, then }\angle \text{PTQ is equal to}\\ \text{ (A) 60° (B) 70°}\\ \text{ (C) 80° (D) 90°}\end{array}$

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Circles

Medium

Q.

$\begin{array}{l}\text{If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, }\\ \text{then }\angle \text{POA is equal to}\\ \text{(A) 50° (B) 60°}\\ \text{(C) 70° (D) 80°}\end{array}$

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Circles

Medium

Q.

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

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Circles

Medium

Q.

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

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Circles

Medium

Q.

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

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Circles

Medium

Q.

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

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Circles

Medium

Q.

$\begin{array}{l}\text{A quadrilateral ABCD is drawn to circumscribe a circle}\\ \text{(see the following figure). Prove that}\\ \mathrm{AB}+\mathrm{CD}=\mathrm{AD}+\mathrm{BC}.\end{array}$

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Circles

Medium

Q.

$\begin{array}{l}\text{In\hspace{0.33em}the\hspace{0.33em}following\hspace{0.33em}figure.\hspace{0.33em}XY\hspace{0.33em}and\hspace{0.33em}X'Y'\hspace{0.33em}are\hspace{0.33em}two\hspace{0.33em}parallel}\\ \text{Tangents\hspace{0.33em}to\hspace{0.33em}a\hspace{0.33em}circle\hspace{0.33em}with\hspace{0.33em}center\hspace{0.33em}O\hspace{0.33em}and\hspace{0.33em}another\hspace{0.33em}}\\ \text{tangent\hspace{0.33em}AB\hspace{0.33em}with\hspace{0.33em}point\hspace{0.33em}of\hspace{0.33em}contact\hspace{0.33em}C\hspace{0.33em}intersecting\hspace{0.33em}XY\hspace{0.33em}at}\\ \text{A\hspace{0.33em}and\hspace{0.33em}X'Y'\hspace{0.33em}at\hspace{0.33em}B.\hspace{0.33em}Prove\hspace{0.33em}that\hspace{0.33em}}\angle {\text{AOB\hspace{0.33em}=\hspace{0.33em}90}}^{\circ }\text{.}\end{array}$

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Circles

Medium

Q.

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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Circles

Medium

Q.

Prove that the parallelogram circumscribing a circle is a rhombus.

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Circles

Medium

Q.

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 8 cm and 6 cm respectively (see the following figure). Find the sides AB and AC.

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Circles

Medium

Q.

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

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• The angle subtended by a chord at the center of a circle.

• Properties of cyclic quadrilaterals (where the sum of opposite angles is 180°).

• Intersecting chords theorem (which relates the product of the lengths of segments of intersecting chords).

These solutions are explained in a step-by-step manner, helping students apply geometric theorems and formulas in a variety of problems, boosting their conceptual clarity and problem-solving abilities for Class 10 exams.

Q1. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the center is 25 cm. The radius of the circle is
(A) 7 cm (B) 12 cm (C) 15 cm (D) 24.5 cm

Difficulty Level: Easy
Known:
(i) Length of tangent from a point Q, i.e. PQ is 24 cm
(ii) Distance of Q from center, i.e. OQ is 25 cm.

Unknown:
Radius of the circle

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

Solution:

$\triangle OPQ$

△OPQ is a right-angled triangle.
By Pythagoras theorem,

$$OQ^2 = OP^2 + PQ^2$$

OQ2=OP2+PQ2

$$25^2 = r^2 + 24^2$$

252=r2+242

$$625 = r^2 + 576$$

625=r2+576

$$r^2 = 625 - 576 = 49$$

r2=625−576=49

$$r = \sqrt{49} = 7$$

r=49​=7

Answer:
Option A

Q2. In the given figure, if TP and TQ are the two tangents to a circle with center O so that ∠POQ = 110°, then ∠PTQ is equal to
(A) 60° (B) 70° (C) 80° (D) 90°

Difficulty Level: Medium
Known:
(i) TP and TQ are tangents to a circle with Center O
(ii) ∠POQ = 110°

Unknown:
∠PTQ

Reasoning:
• Tangent at any point of a circle is perpendicular to the radius through the point of contact.
• In the above figure, OPTQ is a quadrilateral and ∠P and ∠Q are 90°. The sum of the angles of a quadrilateral is 360°.

Solution:
In OPTQ,

$$90^\circ + 90^\circ + 110^\circ + \angle PTQ = 360^\circ$$

90∘+90∘+110∘+∠PTQ=360∘

$$290^\circ + \angle PTQ = 360^\circ$$

290∘+∠PTQ=360∘

$$\angle PTQ = 360^\circ - 290^\circ = 70^\circ$$

∠PTQ=360∘−290∘=70∘

Answer:
Option B

Q3. If tangents PA and PB from a point P to a circle with center O are inclined to each other at an angle of 80°, then ∠POA is equal to
(A) 50° (B) 60° (C) 70° (D) 80°

Difficulty Level: Medium
Known:
PA and PB are the tangents from P to a circle with center O. Tangents are inclined to each other at an angle of 80°

Unknown:
∠POA

Reasoning:
• The lengths of tangents drawn from an external point to a circle are equal.
• Tangent at any point of a circle is perpendicular to the radius through the point of contact.

Solution:
In △OAP and △OBP,
OA = OB (radii of the circle are always equal)
AP = BP (length of the tangents)
OP = OP (common)

By SSS congruency, △OPA ≅ △OBP.
Hence, ∠POA = ∠POB.
Since the angle between PA and PB is 80°,

$$\angle APB = 80^\circ$$

∠APB=80∘

Since OP is the angle bisector of ∠APB,

$$\angle POA = \frac{80^\circ}{2} = 40^\circ$$

∠POA=280∘​=40∘

Answer:
Option A

Q4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Difficulty Level: Medium

Solution:
Given:
• Let AB be a diameter of the circle.
• Two tangents PQ and RS are drawn at points A and B respectively.

To Prove:
Tangents drawn at the ends of a diameter of a circle are parallel.

Reasoning:
• A tangent to a circle is a line that intersects the circle at only one point.
• Theorem 10.1: The tangent at any point of a circle is perpendicular to the radius through the point of contact.

We know that, according to Theorem 10.1, the radius is perpendicular to the tangent at the point of contact.
Thus, OA ⊥ PQ and OB ⊥ RS.
Since the tangents are perpendicular to the radius, ∠PAO = 90° and ∠RBO = 90°.

Here, ∠OAQ = ∠OBR and ∠PAO = ∠OBS are two pairs of alternate interior angles, and they are equal.
By the Alternate Interior Angle Theorem, PQ and RS must be parallel.

Hence, it is proved that tangents drawn at the ends of a diameter of a circle are parallel.

Q5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the center.

Difficulty Level: Medium

To prove:
Perpendicular at the point of contact to the tangent to a circle passes through the center.

Reasoning and Solution:
By Theorem 10.1, tangent at any point of a circle is perpendicular to the radius through the point of contact.
Therefore, the perpendicular PR of tangent PQ passes through center O.

Hence, it is proved that the perpendicular at the point of contact to the tangent to a circle passes through the center.

Q6. The length of a tangent from a point A at distance 5 cm from the center of the circle is 4 cm. Find the radius of the circle.

Difficulty Level: Easy
Known:
Length of tangent from point A = 4 cm.
Distance between the center of the circle and point A is 5 cm.

Unknown:
Radius of the circle

Reasoning:
Tangent at any point of a circle is perpendicular to the radius through the point of contact.
Therefore, ∠OTA = 90° and △OTA is a right-angled triangle.

By Pythagoras theorem,

$$OA^2 = OT^2 + AT^2$$

OA2=OT2+AT2

$$5^2 = r^2 + 4^2$$

52=r2+42

$$25 = r^2 + 16$$

25=r2+16

$$r^2 = 25 - 16 = 9$$

r2=25−16=9

$$r = \sqrt{9} = 3 \, \text{cm}$$

r=9​=3cm

Answer:
Radius of the circle = 3 cm

Q7. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Difficulty Level: Medium
Known:
Two concentric circles are of radii 5 cm and 3 cm.

Unknown:
The length of the chord of the larger circle which touches the smaller circle.

Reasoning:
The chord of the larger circle is a tangent to the smaller circle.
Solution:
PQ is the chord of the larger circle and the tangent of the smaller circle.
Tangent is perpendicular to the radius at the point of contact S.

$\angle OSP = 90^\circ$

∠OSP=90∘

In △OSP (right-angled triangle), by Pythagoras' theorem,

$$OP^2 = OS^2 + SP^2$$

OP2=OS2+SP2

$$5^2 = 3^2 + SP^2$$

52=32+SP2

$$25 = 9 + SP^2$$

25=9+SP2

$$SP^2 = 16 \quad \Rightarrow \quad SP = 4 \, \text{cm}$$

SP2=16⇒SP=4cm

Since the perpendicular from the center bisects the chord, the length of the chord PQ = 2 × SP = 2 × 4 = 8 cm.

Answer:
Length of the chord = 8 cm

Q8. A quadrilateral ABCD is drawn to circumscribe a circle (see given Figure). Prove that:
AB + CD = AD + BC

Difficulty Level: Medium

To prove:
AB + CD = AD + BC

Reasoning:
• A tangent to a circle is a line that intersects the circle at only one point.
• Theorem 10.1: The tangent at any point of a circle is perpendicular to the radius through the point of contact.
• Theorem 10.2: The lengths of tangents drawn from an external point to a circle are equal.

Solution:
We know that the length of tangents drawn from an external point of the circle are equal according to Theorem 10.2.
Therefore,
BP = BQ (Tangents from point B) — (i)
CR = CQ (Tangents from point C) — (ii)
DR = DS (Tangents from point D) — (iii)
AP = AS (Tangents from point A) — (iv)

Adding (i) + (ii) + (iii) + (iv), we get,

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

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

$$AB + CD = AD + BC$$

AB+CD=AD+BC

Hence, proved.

Q9. In Figure, XY and X′Y′ are two parallel tangents to a circle with center O and another tangent AB with point of contact C intersecting XY at A and X′Y′ at B. Prove that ∠AOB = 90°.

Difficulty Level: Medium
Given:
• O is the center of the circle.
• XY and X′Y′ are the two parallel tangents to the circle.
• AB is another tangent with point of contact C, intersecting XY at A and X′Y′ at B.

To prove:
∠AOB = 90°

Reasoning and Solution:
Join point O to C.
In △OPA and △OCA,
OP = OC (Radii of the circle are equal)
AP = AC (The lengths of tangents drawn from an external point A to a circle are equal.)
AO = AO (Common)

By SSS congruency, △OPA ≅ △OCA.
Therefore, ∠POA = ∠AOC.

Similarly, △OQB ≅ △OCB.
Therefore, ∠BOQ = ∠QOC.

Since PQ is a diameter, hence a straight line and
∠POQ = 180°.
But

$$\angle POQ = \angle POA + \angle AOB + \angle BOQ$$

∠POQ=∠POA+∠AOB+∠BOQ

$$\angle AOB = 90°$$

∠AOB=90°

Hence, proved.

Q10. 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 center.

Difficulty Level: Medium
Given:
• Let us consider a circle centered at point O.
• Let P be an external point from which two tangents PA and PB are drawn to the circle which are touching the circle at point A and B respectively.
• AB is the line segment, joining point of contacts A and B together such that it subtends ∠AOB at center O of the circle.

To prove:
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 point of contact at the center.
i.e. ∠APB is supplementary to ∠AOB

Reasoning and Solution:
According to Theorem 10.1: The tangent at any point of a circle is perpendicular to the radius through the point of contact.

$$\angle OAP = \angle OBP = 90^\circ \quad \text{(Equation 1)}$$

∠OAP=∠OBP=90∘(Equation 1)

In a quadrilateral, sum of 4 angles is 360°. Hence,

$$\angle OAP + \angle APB + \angle PBO + \angle BOA = 360°$$

∠OAP+∠APB+∠PBO+∠BOA=360°

Substituting the values from Equation 1, we get,

$$90^\circ + \angle APB + 90^\circ + \angle BOA = 360°$$

90∘+∠APB+90∘+∠BOA=360°

$$180^\circ + \angle APB + \angle BOA = 360°$$

180∘+∠APB+∠BOA=360°

$$\angle APB + \angle BOA = 180^\circ$$

∠APB+∠BOA=180∘

Hence, proved.

Q11. Prove that the parallelogram circumscribing a circle is a rhombus.

Difficulty Level: Medium

To prove:
The parallelogram circumscribing a circle is a rhombus.

Reasoning and Solution:
ABCD is a parallelogram. Therefore, opposite sides are equal.

$$AB = CD \quad \text{and} \quad BC = AD$$

AB=CDandBC=AD

According to Theorem 10.2: The lengths of tangents drawn from an external point to a circle are equal.
Therefore,
BP = BQ (Tangents from point B) — (1)
CR = CQ (Tangents from point C) — (2)
DR = DS (Tangents from point D) — (3)
AP = AS (Tangents from point A) — (4)

Adding (1) + (2) + (3) + (4), we get,

$$AB + CD = AD + BC$$

AB+CD=AD+BC

Since

$AB = BC$

AB=BC and

$CD = AD$

CD=AD, all sides are equal.

Therefore, the parallelogram circumscribing a circle is a rhombus.

Q12. 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 8 cm and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC.

Difficulty Level: Hard
Unknown:
Sides AB and AC

Known:
(i) Triangle ABC is drawn to circumscribe a circle of radius 4 cm.
BD = 8 cm
CD = 6 cm

Reasoning:
Finding the area of triangle ABC in two ways and equating them will result in unknown length. Hence other sides can be calculated.
Let
AE = AF = x (The lengths of tangents drawn from an external point to a circle are equal.)
CD = CE = 6 cm (Tangents from point C)
BD = BF = 8 cm (Tangents from point B)

Area of the triangle =

$\sqrt{s(s-a)(s-b)(s-c)}$

s(s−a)(s−b)(s−c)​ where

$s = \frac{a+b+c}{2}$

s=2a+b+c​

By calculating the areas using different methods and solving, we get:
AB = 15 cm
AC = 13 cm

Answer:
AB = 15 cm
AC = 13 cm

FAQs: Class 10 Maths Chapter 10 – Circles Exercise 10.2

Q1. What is the main focus of Exercise 10.2?
Answer:
Exercise 10.2 focuses on the properties of chords and angles subtended by them at the center, including cyclic quadrilaterals and intersecting chords.

Q2. What is the angle subtended by a chord at the center of a circle?
Answer:
The angle subtended by a chord at the center of a circle is twice the angle subtended at any point on the remaining part of the circle (i.e., the angle at the center is twice the angle at the circumference).

Q3. What is a cyclic quadrilateral?
Answer:
A cyclic quadrilateral is a quadrilateral whose vertices lie on a circle. The key property of cyclic quadrilaterals is that the sum of the opposite angles is always 180°.

Q4. What is the intersecting chords theorem?
Answer:
The intersecting chords theorem states that if two chords AB and CD intersect at a point

$P$

P, then the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other chord:

$$PA \times PB = PC \times PD$$

PA×PB=PC×PD

Q5. How do NCERT Solutions help with exam preparation?
Answer:
These solutions provide clear explanations of theorems and their applications in real problems, helping students apply concepts effectively. Regular practice with these solutions strengthens problem-solving skills and improves exam performance.