Class 10 NCERT Solutions
Chapter 10: Circles
Master the properties of tangents, the theorems of lengths from external points, and the geometric logic of circular contact with our step-by-step proof-based approach.
Exercise 10.1
1. How many tangents can a circle have?
A circle can have infinitely many tangents, since there are infinitely many points on a circle and a unique tangent can be drawn at each point.
2. Fill in the blanks:
(i) A tangent to a circle intersects it in _____ point(s).
(ii) A line intersecting a circle in two points is called a _____.
(iii) A circle can have _____ parallel tangents at the most.
(iv) The common point of a tangent to a circle and the circle is called _____.
(i) one
(ii) secant
(iii) two
(iv) point of contact
3. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is:
(A) 12 cm (B) 13 cm (C) 8.5 cm (D) $\sqrt{119}$ cm
Since OP $\perp$ PQ, we apply the Pythagorean theorem in $\triangle\text{OPQ}$:
$$\text{PQ} = \sqrt{\text{OQ}^2-\text{OP}^2} = \sqrt{12^2-5^2} = \sqrt{144-25} = \sqrt{119} \text{ cm}$$Answer: (D)
Exercise 10.2
1. 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
Using the Pythagorean theorem (radius $\perp$ tangent):
$$r = \sqrt{25^2-24^2} = \sqrt{625-576} = \sqrt{49} = 7 \text{ cm}$$Answer: (A)
2. In figure, 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) $90°$
Since TQ and TP are tangents, $\text{OP} \perp \text{PT}$ and $\text{OQ} \perp \text{QT}$, giving $\angle\text{OPT} = \angle\text{OQT} = 90°$.
In quadrilateral TPOQ, the angle sum is $360°$:
$$\angle\text{PTQ} + 90° + 110° + 90° = 360° \Rightarrow \angle\text{PTQ} = 70°$$Answer: (B)
3. 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) $80°$
The angle $\angle\text{APB} = 80°$, so $\angle\text{AOB} = 180° – 80° = 100°$ (supplementary angles in quadrilateral OAPB).
$$\angle\text{POA} = \frac{1}{2}\angle\text{AOB} = \frac{1}{2}\times100° = 50°$$Answer: (A)
4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Let PQ be the diameter and let tangents AB (at P) and CD (at Q) be drawn.
Since a tangent is perpendicular to the radius at the point of contact:
$$\text{PQ} \perp \text{AB} \Rightarrow \angle\text{APQ} = 90°$$ $$\text{PQ} \perp \text{CD} \Rightarrow \angle\text{PQD} = 90°$$Since $\angle\text{APQ} = \angle\text{PQD} = 90°$, these are alternate interior angles made by transversal PQ.
$\therefore \text{AB} \| \text{CD}$. $\blacksquare$
5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Suppose the perpendicular at point of contact T does NOT pass through the centre O. Let OP be this supposed perpendicular.
Since OP is the radius, $\text{OP} \perp \text{PT}$ (radius is perpendicular to tangent at point of contact).
But we assumed OP is also perpendicular to PT. Since only one perpendicular can be drawn from a point to a line, OP and the radius OP must coincide.
Therefore, the perpendicular at the point of contact necessarily passes through the centre. $\blacksquare$
6. 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.
Using the Pythagorean theorem (radius $\perp$ tangent):
$$r = \sqrt{5^2-4^2} = \sqrt{25-16} = \sqrt{9} = 3 \text{ cm}$$
7. 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.
$\text{OA} = 5$ cm, $\text{OT} = 3$ cm, and $\text{OT} \perp \text{AB}$. Using Pythagoras:
$$\text{AT} = \sqrt{25-9} = \sqrt{16} = 4 \text{ cm}$$ $$\therefore \text{AB} = 2\,\text{AT} = 8 \text{ cm}$$
8. A quadrilateral ABCD is drawn to circumscribe a circle. Prove that $AB + CD = AD + BC$.
Using the property that tangents drawn from an external point to a circle are equal:
$$\text{AS} = \text{AP},\quad \text{BP} = \text{BQ},\quad \text{CQ} = \text{CR},\quad \text{DR} = \text{DS}$$ $$\text{AB}+\text{CD} = (\text{AP}+\text{PB})+(\text{CR}+\text{RD}) = \text{AS}+\text{BQ}+\text{CQ}+\text{DS}$$ $$= (\text{AS}+\text{DS})+(\text{BQ}+\text{CQ}) = \text{AD}+\text{BC} \qquad \blacksquare$$
9. 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 $\angle AOB = 90°$.
In $\triangle\text{AOP}$ and $\triangle\text{AOC}$: PA = PC, PO = CO (radii), AO = AO (common). By SSS, $\triangle\text{AOP} \cong \triangle\text{AOC}$, so $\angle 1 = \angle 2$.
Similarly, $\angle 3 = \angle 4$.
Since XY $\|$ X’Y’, the co-interior angles satisfy:
$$\angle\text{PAB}+\angle\text{QBA} = 180° \Rightarrow 2\angle 2+2\angle 3 = 180° \Rightarrow \angle 2+\angle 3 = 90°$$ $$\therefore \angle\text{AOB} = 90° \qquad \blacksquare$$
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.
Since OA $\perp$ PA and OB $\perp$ PB, we have $\angle\text{OAP} = \angle\text{OBP} = 90°$.
In quadrilateral OAPB, the angle sum equals $360°$:
$$\angle\text{AOB}+90°+\angle\text{BPA}+90° = 360° \Rightarrow \angle\text{AOB}+\angle\text{BPA} = 180° \qquad \blacksquare$$
11. Prove that the parallelogram circumscribing a circle is a rhombus.
For a parallelogram, $\text{AB} = \text{CD}$ and $\text{BC} = \text{DA}$. Using tangent-length equality:
$$\text{AB}+\text{CD} = \text{AP}+\text{BP}+\text{CR}+\text{DR} = \text{AS}+\text{BQ}+\text{CQ}+\text{DS} = \text{AD}+\text{BC}$$ $$\Rightarrow 2\text{AB} = 2\text{AD} \Rightarrow \text{AB} = \text{AD}$$Since all adjacent sides of the parallelogram are equal, it is a rhombus. $\blacksquare$
12. 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. Find the sides AB and AC.
Let the circle touch AB at E and AC at F, with $\text{AE} = \text{AF} = x$ cm.
Also $\text{CD} = 6 = \text{CF}$, $\text{BD} = 8 = \text{BE}$. The semi-perimeter is $s = 14+x$.
Equating the area using Heron’s formula and the formula with inradius:
$$\sqrt{(14+x)\cdot6\cdot x\cdot8} = \sqrt{48x(14+x)} = (16+2x)+28+(12+2x) = 4(14+x)$$Squaring: $48x(14+x) = 16(14+x)^2 \Rightarrow 48x = 16(14+x) \Rightarrow 2x = 14 \Rightarrow x = 7$ cm
$$\text{AB} = 8+7 = \mathbf{15} \text{ cm}, \quad \text{AC} = 6+7 = \mathbf{13} \text{ cm}$$
13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Using congruent triangles at each vertex (similar to Q.8 approach), we establish:
$$\angle 1 = \angle 8,\quad \angle 2 = \angle 3,\quad \angle 4 = \angle 5,\quad \angle 6 = \angle 7$$Since the total of all eight angles around centre O is $360°$:
$$2\angle 1+2\angle 2+2\angle 6+2\angle 5 = 360°$$ $$\Rightarrow (\angle 1+\angle 2)+(\angle 5+\angle 6) = 180°$$ $$\Rightarrow \angle\text{AOB}+\angle\text{COD} = 180°$$Similarly, $\angle\text{AOD}+\angle\text{BOC} = 180°$. $\blacksquare$
