CIRCLES-Exercise 10.2

The Solutions to the Textbook Exercises of NCERT Mathematics Class X Chapter 10 – “Circles” are designed to provide students with clear, logically structured, and exam-ready explanations for every problem prescribed in the syllabus. This chapter plays a crucial role in strengthening deductive reasoning, as most questions require a sound understanding of definitions, theorems, and their correct application through well-drawn diagrams and step-wise proofs. These solutions focus on developing conceptual clarity by explaining why each step is taken, rather than merely presenting final answers. Special emphasis is placed on proof-based questions involving tangents, perpendicularity of radius, equality of tangents from an external point, and construction-oriented problems that demand precision and accuracy. Each solution follows the board-preferred format, ensuring proper mathematical language, neat presentation, and logical flow, which are essential for scoring full marks in examinations. By practising these solved textbook exercises thoroughly, students gain confidence in handling a wide variety of questions—ranging from short answers to long proofs—while also improving their diagram interpretation and reasoning skills. This solution set serves as a reliable companion for revision, self-assessment, and last-minute exam preparation, fully aligned with the expectations of Class X board examinations

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December 14, 2025  |  By Academia Aeternum

CIRCLES-Exercise 10.2

Maths - Exercise
Fig. 10.2.1-x
Fig. 10.2.1

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

Solution

Length of tangent from point Q = 24 m
Distance of the point Q from the centre of the circle = 25 cm
The radius of the circle OP can be calculated by pythagoras theorem

$$\begin{aligned}OQ^{2}&=OQ^{2}-PQ^{2}\\ OP^{2}&=25^{2}-24^{2}\\ &=625-576\\ &=49\\ OP&=\sqrt{49}\\ &=7cm\end{aligned}$$ Hence
Option (A) 7 cm is correct
Fig. 10.11-x
Fig. 10.11

Q2. In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that \(\angle POQ = 110°\), then \(\angle PTQ\) is equal to
(A) 60°
(B) 70°
(C) 80°
(D) 90°

Solution

TP and TQ are tangents \(\angle POQ = 110°\)
\(angle P=\angle Q =90^\circ) (Angles subtended by radius on tangents)

$$\begin{aligned}\angle O&=110^{\circ}\\ \angle P&=90^{\circ}\\ \angle Q&=90^{\circ }\\ \angle O+\angle P+\angle Q+\angle T&=360^{\circ}\\ \scriptsize\text{(Sum of internal angles of a }&\scriptsize\text{quadrilateral)}\\\\ 110^{\circ}+90^{\circ}+90^{\circ}+\angle T&=360^{\circ}\\ 290^{\circ }+\angle T&=360^{\circ }\\ \angle T&=360-20^{\circ }\\ &=70^{\circ}\\ \angle PTQ&=70^{\circ }\end{aligned}$$

Hence,
Option (B) 70° is correct

Fig. 10.2.3-x
Fig. 10.2.3

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

Solution

PA and PB are tangents
OA is the angle bisector of the tangents

In \(\triangle PAO and \triangle PBO\)

\[ \begin{aligned} AO&=AO\\ \scriptsize\text{(Common Hy}&\scriptsize\text{potenuse)}\\ \angle PAO&=\angle PBO\\ \scriptsize\text{(angle subtended by}&\scriptsize\text{ radius of tangent)}\\ AO&=AB\quad\scriptsize\text{(radii)}\\ \therefore \triangle PAO &\cong \triangle PBO\quad\scriptsize\text{(RHS)}\\ \angle APO&=\angle BPO\quad\scriptsize\text{(CPCT)} \end{aligned} \]

therefore angle PBO = 80/2 = 40°

$$\scriptsize\begin{aligned}\angle DOA+\angle OAP+\angle APO&=180^{\circ}\\ \scriptsize\text{(Angle Sum Property)}\\\\ \angle POA+90^{\circ}+40^{\circ }&=180^{\circ }\\ \angle POA&=180^{\circ}-130^{\circ }\\ &=50^{\circ }\end{aligned}$$

Hence,
Option (A) 50° is correct

Fig. 10.2.4-x
Fig. 10.2.4

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

Solution

To prove:
tangents drawn at the end of diameter are parallel

construction: Draw tangents PQ and RS at the end of the diameter of the circle

Proof:

Let AB is diameter of the circle and
O is the center of the circle

$$\begin{aligned}\angle PAO&=90^{\circ }\\ \angle SBO&=90^{0}\end{aligned}$$ \(\angle PAO\) and \(\angle SBO\) are Alternate angle to the transversal AB
Hence, $$ PQ\parallel RS$$ Thus Proved.
Fig. 10.2.5-x
Fig. 10.2.5

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

Solution

Given:
A circle with centre O and a tangent AB touching the circle at point P

To Prove:
The line drawn through P and perpendicular to AB passes through the centre O

Construction:
Join OP, the radius from the centre O to the point of contact P.

Proof:
  • From the basic tangent theorem, the tangent at any point of a circle is perpendicular to the radius through the point of contact; that is, \[OP\perp AB\]
  • At a given point on a line (here, pointP on tangent AB), there is exactly one unique line that is perpendicular to that line.
  • The line through P, which is perpendicular to AB is therefore the same line as OP
  • Since OP joins the point of contact P to the centre O, this perpendicular must pass through the centre of the circle

Hence, the perpendicular at the point of contact to the tangent to a circle passes through the centre of the circle.
Fig. 10.2.6-x
Fig. 10.2.6

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

Solution

Length of tangent from point A = 4cm
Length of the point A from the centre of the Circle = 5cm
The radius of the circle can be calculated using pythagoras theorem

$$\begin{aligned}AO^{2}&=AB^{2}+OB^{2}\\ 5^{2}&=4^{2}+OB^{2}\\ OB^{2}&=5^{2}-4^{2}\\ &=25-16\\ &=9\\ OB&=\sqrt{9}\\ OB&=3\end{aligned}$$ Radius of the circle = 3cm
Fig. 10.2.7-x
Fig. 10.2.7

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.

Solution

Radius of Concentric circle = 5cm & 3cm
chord that touches in her circle is also tangent to the inner circle
Hence angle subtended by the radios on the chord is 90°
If the radius subtends 90° on a chord, then it bisects the chord calculating half of the chord by Pythagoras' theorem $$\begin{aligned}OR^{2}+PR^{2}&=OP^{2}\\ PR^{2}&=OP^{2}-OR^{2}\\ &=5^{2}-3^{2}\\ &=25-9\\ &=16\\ \Rightarrow PR&=\sqrt{16}\\ &=4\end{aligned}$$

Hence length of chord \(PQ = 2 PR \Rightarrow 2\times 4 = 8\) cm

Fig. 10.12-x
Fig. 10.12

Q8. A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that
AB + CD = AD + BC

Solution

8. Given: \(ABCD\) is a quadrilateral
To Prove: \[AB + CD = AD + BC\] Proof: \[AP = AS\tag{1}\] (target drawn from point A) \[BP = BQ\tag{2}\] (tangent drawn from point B) \[CR = CQ\tag{3}\] (target drawn from point C) \[DR = DS \tag{4}\] (tanged drawn from ppoit D) Adding (1) and (2) \[\begin{align}AP + B P &= AS + BQ\\ \Rightarrow AB &= AS + BQ\tag{4}\end{align}\] Adding (3) and (4) \[\begin{align}CR + DR &= CQ + DS\\ \Rightarrow CD &= CQ + DS\tag{5}\end{align}\] Adding (5) and (6) \[\scriptsize\begin{align}AB + CD &= AS + BQ + CQ + DS\\ &= AS + DS + BQ + CQ\\ &= AD + BC\\ AB + C &= AD + BC\end{align}\] Hence Proved
Fig. 10.13-x
Fig. 10.13

Q9. In Fig. 10.13, 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 \(\angle AOB = 90°\).

Solution

$$\angle PAB+\angle ABQ=180^{0}\tag{1} $$

sum of angles on same side of transversal AB

$$\begin{aligned}\angle PAO&=\angle BAO\\ \angle PAB&=\angle PAO+\angle BAO\\ \angle PAB&=2BAO\end{aligned}$$ $$\begin{aligned}\angle QBO&=\angle OBA\\ \angle ABQ&=\angle QBO+\angle OBA\\ &=2\angle OBA\end{aligned}$$

From Equation-(1)

$$\begin{aligned}\angle PAB+\angle ABQ&=180^{\circ }\\ 2\angle BAO+2\angle OBA&=180^{0}\\ 2\left( \angle BAO+\angle OBA\right) &=180^{\circ }\\ \angle BAO+\angle OBA&=90^{0}\end{aligned}$$

But, by Angle sum property

$$\scriptsize\begin{aligned}\angle AOB+\angle BAO+\angle OBA&=180^{\circ }\\ \angle AOB+\left( \angle BAO+\angle OBA\right) &=180^{\circ }\\ \angle AOB+90^{\circ }&=180^{\circ }\\ \angle AOB&=180^{\circ }-90^{\circ }\\ \angle AOB&=90^{\circ }\end{aligned}$$ Hence Proved
Fig. 10.2.10-x
Fig. 10.2.10

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 centre.

Solution

Given:

AB and AC are targets drawn from point A


To Prove:
\(\angle BOC + \angle BAC = 180°\)

Proof:

Sum of internal angles of a quadrilateral is = 360°
\(\angle OBA = 90°\) (Angle between tangent AB and radius OB)
\(\angle OCA = 90°\) (angle between tangent Ac and radius OC)

$$\scriptsize\begin{aligned}\angle ABO+\angle BOC+\angle OCA+\angle CAB&=360^{\circ}\\ \angle BOC+\angle CAB+90^{\circ}+90^{\circ}&=360^{\circ }\\ \angle BOC+\angle CAB&=360^{\circ }-180^{\circ }\\ \angle BOC+\angle CAB&=180^{\circ}\end{aligned}$$ Hence, angles BOC and angle CAB are supplementary angles
Fig. 10.2.11-x
Fig. 10.2.11

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

Solution

Proposition:
A parallelogram that circumscribes a circle is a rhombus.

Proof:
Let ABCD be a parallelogram that circumscribes a circle.
Let the circle touch AB, BC, CD, DA at P, Q, R, S respectively.
From the property of tangents drawn from an external point to a circle, the lengths of the two tangents from the same point are equal. Hence, \[ \begin{aligned} AP &= AS,\\ BP &= BQ,\\ CQ &= CR,\\ DR &= DS \end{aligned} \] Now, the sides of the parallelogram can be expressed as: \[ \begin{aligned} AB &= AP + PB\\ BC &= BQ + QC\\ CD &= CR + RD\\ DA &= DS + SA \end{aligned} \] Substituting the equal tangent segments, we get \[ \begin{aligned} AB &= AP + PB\\ BC &= PB + AP\\ CD &= AP + PB\\ DA &= AP + PB \end{aligned} \] Thus, \[ AB = BC = CD = DA. \] Since all four sides of the parallelogram are equal, the parallelogram is a rhombus.

Hence, a parallelogram circumscribing a circle is a rhombus.
Fig. 10.2.12-x
Fig. 10.2.12

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

Solution

\(BD = BE = 8\) (tangent from point \(B\))
\(CD = CF = 6\) (tangent from point C)
Let \(AE = AF = x\) (tangent from Point \(A\)) Perimeter of triangle ABC =

$$\begin{aligned}p&=6+8+8+x+x+6\\ &=2x+28\end{aligned}$$ Semi Perimeter (s) $$\begin{aligned}S&=\dfrac{P}{2}&=\dfrac{1}{2}\left( 2x+28\right) \\ &=x+14\end{aligned}$$

Area of \(\triangle ABC\) = Area of \(\triangle AOC\) + Area of \(\triangle COB\) + Area of \(\triangle BOA\)

Area of \(\triangle ABC\) (Heron's Formula)

$$\scriptsize\begin{aligned}\Delta &=\sqrt{s\left( s-a\right) \left( s-b\right) \left( s-c\right) }\\ &=\sqrt{\left( x+14\right) \left( x+14-14\right) ( x+14-\left( 8+x\right) ( x+14-\left( 6+x\right) }\\ &=\sqrt{\left( x+14\right) \left( x\right) \left( 6\right) \left( 8\right) }\\ &=\sqrt{48x\left( x+14\right) }\end{aligned}$$

Area of \(\triangle AOC\) Base \(b\)= x+6
Height \(h\)= radius = 4 (Radius is perpendicular on Tangent)

$$\begin{aligned} ar(\triangle AOC)&=\dfrac{1}{2}\times b\times h\\\\\ &=\dfrac{1}{2}\left( x+6\right) \times 4\\\\ &=2\left( x+6\right) \\ &=2x+12\end{aligned}$$

\(\triangle COB\) Base \(b\) = 14,
Height \(h\)= radius = 4 (Radius is perpendicular on Tangent)

$$\begin{aligned} ar(\triangle AOC)&=\dfrac{1}{2}\times b\times h\\\\\ &=\dfrac{1}{2}\times 14\times 4\\ &=28\end{aligned}$$

Area of \(\triangle BOA\) Base \(b = x+8\),
Height \(h\)= radius = 4 (Radius is perpendicular on Tangent)

$$\begin{aligned} ar(\triangle BOA)&=\dfrac{1}{2}\times b\times h\\\\\ &=\dfrac{1}{2}\cdot \left( x+8\right) \times 4\\ &=2x+16\end{aligned}$$

Sum of the areas of inner triangles

$$\begin{aligned}&2x+12+28+x+16\\ &=4x+56\end{aligned}$$

Area of \(\triangle\) ABC = Sum of inner triangles

$$\begin{aligned}\sqrt{48x\left( x+14\right) }&=4x+56\\ \scriptsize\text{Squaring }&\scriptsize\text{Both Sides}\\ 48x\left( x+14\right) &=\left[ 4\left( x+14\right) \right] ^{2}\\ 48x\left( x+14\right) &=16\left( x+14\right) ^{2}\\ 3x\left( x+14\right) &=\left( x+14\right) ^{2}\\ 3x&=\dfrac{\left( x+14\right) ^{2}}{x+14}\\ 3x&=x+14\\ 2x&=14\\ x&=\dfrac{14}{2}\\ x&=7cm\end{aligned}$$ Therefore side AC = 7+6 = 13 cm side AB = 7+8 = 15 cm
Fig. 10.2.13-x
Fig. 10.2.13

Q13. Prove that the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Solution

Proposition:
Opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Proof:
Let ABCD be a quadrilateral circumscribing a circle with centre O.
Let the circle touch sides AB, BC, CD, DA at P, Q, R, S respectively.

Join OP, OQ, OR, OS.

Since a radius drawn to the point of contact is perpendicular to the tangent, \[ \begin{aligned} OP &\perp AB\\ OQ &\perp BC\\ OR &\perp CD\\ OS &\perp DA \end{aligned} \]

Consider the angles subtended at the centre by opposite sides AB and CD.
The angle subtended by side AB at the centre is \(\angle POS\),
and the angle subtended by side CD at the centre is \(\angle QOR.\)

In quadrilateral \(POSQ,\)
angles at P and Q are right angles. Hence,

\[ \angle POS + \angle QOR = 180^\circ. \]

Similarly, the angles subtended by the other pair of opposite sides BC and DA at the centre are also supplementary.

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

Frequently Asked Questions

A circle is the set of all points in a plane that are at a fixed distance from a fixed point called the centre.

The centre is the fixed point from which all points on the circle are equidistant.

The radius is the line segment joining the centre of the circle to any point on its circumference.

The diameter is a chord passing through the centre of the circle and is twice the radius.

A chord is a line segment joining any two points on the circumference of a circle.

The diameter is the longest chord of a circle.

A secant is a line that intersects a circle at two distinct points.

A tangent is a line that touches a circle at exactly one point.

The point where a tangent touches the circle is called the point of contact.

A tangent has exactly one common point with the circle.

A secant has exactly two common points with the circle.

The tangent at any point of a circle is perpendicular to the radius through the point of contact.

The angle is always a right angle (90°).

Exactly one tangent can be drawn from a point on the circle.

No tangent can be drawn from a point inside the circle.

Two tangents can be drawn from a point outside the circle.

The lengths of tangents drawn from an external point to a circle are equal.

They form congruent right-angled triangles with equal radii and a common hypotenuse.

No, a tangent always lies outside the circle except at the point of contact.

It is the square of the length of the tangent drawn from the point to the circle.

The radius drawn to the point of contact is perpendicular to the tangent, not the chord.

No, a diameter always passes through the interior of the circle.

No, a chord is only the segment inside the circle, while a secant is the entire line.

No, but every secant contains a chord within the circle.

The chord formed becomes the diameter of the circle.

The angle between a tangent and a chord equals the angle in the opposite arc of the circle.

It is used to find angles formed by tangents and chords in a circle.

Right-angled triangles are formed with the radius perpendicular to the tangent.

The perpendicular distance from the centre to the tangent is the shortest.

Because the perpendicular gives the shortest distance from the centre to the tangent.

No, they meet at the external point and touch the circle at different points.

The internal segment lies inside the circle; the external segment lies outside.

Constructing tangents from an external point and at a point on the circle.

Yes, in wheels, circular tracks, mechanical parts, and optical instruments.

They help establish right angles, congruence, and length relationships.

Yes, tangents help analyze symmetry and rotational properties of circles.

The tangent–radius perpendicularity theorem.

Proofs, constructions, angle finding, and length-based problems.

Yes, especially based on equal tangents from an external point.

Because properties of tangents and secants are visual and diagram-dependent.

Yes, one tangent at each point on the circle.

It is the path traced by a point moving at a constant distance from a fixed point.

It connects geometry, constructions, and logical proofs.

No, conceptual clarity and diagram understanding are essential.

Learn theorems, practice diagrams, and write step-wise proofs clearly.

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