Geometry plays a key role in understanding shapes, symmetry, and patterns we see in everyday life. NCERT Solutions for Class 10 Maths Chapter 10, Circles, helps students master the core properties of circles, with a strong focus on tangents, radii, chords, and theorems that are frequently asked in exams.
This chapter explains important concepts like tangent to a circle, number of tangents from an external point, and key results such as “tangent is perpendicular to the radius at the point of contact” and “lengths of tangents from an external point are equal”. Chapter 10 is highly important for the CBSE Class 10 board exams and also forms a foundation for
advanced geometry topics in higher classes.
NCERT Solutions for Class 10 Maths Chapter 10 Circles
Q.
How many tangents can a circle have?
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.
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.
Q.
Prove that the parallelogram circumscribing a circle is a rhombus.
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.
Q.
Q.
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.
Q.
Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Q.
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 _______.
Q.
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Q.
Q.
Q.
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
Q.
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.
Q.
Q.
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Q1. How many tangents can a circle have?
Solution:
A circle can have infinitely many tangents because a tangent can be drawn at every point on the circle, and a circle has infinitely many points.
Q2. 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 _______.
Solution:
(i) one
(ii) secant
(iii) two
(iv) point of contact
Q3. 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:
Solution:
Radius OP is perpendicular to tangent PQ, so ∠OPQ = 90°.
In right △OPQ, OQ = 12 cm and OP = 5 cm.
Using Pythagoras theorem:
OQ2 = OP2 + PQ2
122 = 52 + PQ2
144 = 25 + PQ2
PQ2 = 119
PQ = √119 cm
So, the correct option is (D) √119 cm.
Q4. 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.
Solution (Construction):
1) Draw a circle with centre O and radius r.
2) Draw a line l (given line).
3) Draw a line l1 parallel to l such that its distance from O is exactly r.
→ Then l1 will touch the circle at one point, so it is a tangent.
4) Draw another line l2 parallel to l such that its distance from O is less than r.
→ Then l2 will cut the circle at two points, so it is a secant.
Q5. 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:
Solution:
Let O be the centre and QT be tangent touching circle at T.
OT ⟂ QT, so △OQT is right-angled at T.
Given: OQ = 25 cm, QT = 24 cm, OT = r
Using Pythagoras theorem:
OQ2 = OT2 + QT2
252 = r2 + 242
625 = r2 + 576
r2 = 49
r = 7 cm
So, the correct option is (A) 7 cm.
Q6. If TP and TQ are two tangents to a circle with centre O and ∠POQ = 110°, then ∠PTQ is equal to:
Solution:
We know: Angle between two tangents from an external point is supplementary to the angle subtended at the centre by the chord joining the points of contact.
So, ∠PTQ + ∠POQ = 180°
∠PTQ = 180° − 110° = 70°
Correct option: (B) 70°
Q7. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80°, then ∠POA is equal to:
Solution:
Angle between tangents: ∠APB = 80°
We know: ∠APB + ∠AOB = 180°
So, ∠AOB = 180° − 80° = 100°
Also, OP bisects ∠AOB (since OA = OB radii, symmetry about OP).
Therefore, ∠POA = ½ ∠AOB = ½ × 100° = 50°
Correct option: (A) 50°
Q8. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Solution:
Let AB be a diameter of the circle with centre O.
Tangents are drawn at A and B, say l1 at A and l2 at B.
We know: Tangent at a point is perpendicular to the radius at that point.
So, l1 ⟂ OA and l2 ⟂ OB.
But OA and OB lie on the same straight line AB (diameter).
Hence both l1 and l2 are perpendicular to the same line AB.
Therefore, l1 ∥ l2 (parallel). Proved.
Q9. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Solution:
Let a circle have centre O and tangent l touches it at point P.
We know: The tangent at P is perpendicular to the radius OP.
So, OP ⟂ l at P.
Hence the line drawn perpendicular to the tangent at P is the same as OP, which passes through O (the centre).
Therefore, the perpendicular at the point of contact passes through the centre. Proved.
Q10. 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.
Solution:
Let O be the centre and AT be tangent at T.
OT ⟂ AT, so △OAT is right-angled at T.
Given: OA = 5 cm, AT = 4 cm, OT = r
Using Pythagoras theorem:
OA2 = OT2 + AT2
52 = r2 + 42
25 = r2 + 16
r2 = 9
r = 3 cm
Q11. 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:
Let O be the common centre. Larger radius R = 5 cm, smaller radius r = 3 cm.
Chord of larger circle touches smaller circle ⇒ distance of chord from centre O is 3 cm.
Let chord length = 2x. Perpendicular from centre to chord bisects it.
So, in right triangle: (half chord)² + (distance from centre)² = R²
x2 + 32 = 52
x2 + 9 = 25
x2 = 16 ⇒ x = 4
Chord length = 2x = 8 cm
Q12. A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC.
Solution:
Let the circle touch AB, BC, CD and DA at P, Q, R and S respectively.
From an external point, tangents drawn to a circle are equal.
So, from A: AP = AS
From B: BP = BQ
From C: CQ = CR
From D: DR = DS
Now,
AB = AP + PB
BC = BQ + QC
CD = CR + RD
AD = AS + SD
Add AB + CD:
AB + CD = (AP + PB) + (CR + RD)
= (AS + BQ) + (CQ + SD) (substituting equals)
= (AS + SD) + (BQ + QC)
= AD + BC
Hence, AB + CD = AD + BC. Proved.
Q13. In the following 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 ∠AOB = 90°.
Solution:
Since XY and X'Y' are tangents, radius to each point of contact is perpendicular to the tangent.
Also, AB is tangent at C, so OC ⟂ AB.
Using the property of angles formed by tangents and radii, and the fact that tangents from A and B create right angles with corresponding radii, we get that OA and OB are along the angle bisector directions formed by perpendicular tangents.
Hence, the angle between OA and OB becomes a right angle.
Therefore, ∠AOB = 90°. Proved.
Q14. 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:
Let PA and PB be tangents from external point P to a circle with centre O, touching at A and B.
OA ⟂ PA and OB ⟂ PB ⇒ ∠OAP = 90° and ∠OBP = 90°.
In quadrilateral OAPB:
∠OAP + ∠OBP + ∠AOB + ∠APB = 360°
90° + 90° + ∠AOB + ∠APB = 360°
∠AOB + ∠APB = 180°
Thus, angle between tangents (∠APB) is supplementary to angle at centre (∠AOB). Proved.
Q15. Prove that the parallelogram circumscribing a circle is a rhombus.
Solution:
Let parallelogram ABCD circumscribe a circle.
For any tangential quadrilateral: AB + CD = BC + AD.
But in a parallelogram: AB = CD and BC = AD.
So, AB + AB = BC + BC ⇒ 2AB = 2BC ⇒ AB = BC.
Thus adjacent sides are equal, and with opposite sides already equal, all four sides are equal.
Therefore, ABCD is a rhombus. Proved.
Q16. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that BD = 8 cm and DC = 6 cm. Find sides AB and AC.
Solution:
Let the circle touch AB, BC, CA at E, D, F respectively.
Tangents from same external point are equal:
From B: BE = BD = 8
From C: CF = CD = 6
Let AE = AF = x
Then,
AB = AE + BE = x + 8
AC = AF + CF = x + 6
BC = BD + DC = 8 + 6 = 14
Now semiperimeter s = (AB + BC + AC)/2 = (x+8 + 14 + x+6)/2 = (2x+28)/2 = x+14
Area of triangle = r × s = 4(x+14)
Also, using Heron’s formula with sides (x+8), 14, (x+6):
Area = √[s(s-a)(s-b)(s-c)]
= √[(x+14){(x+14)-(x+8)}{(x+14)-14}{(x+14)-(x+6)}]
= √[(x+14)(6)(x)(8)]
= √[48x(x+14)]
Equate areas:
4(x+14) = √[48x(x+14)]
Square both sides:
16(x+14)2 = 48x(x+14)
Divide by (x+14):
16(x+14) = 48x
16x + 224 = 48x
224 = 32x
x = 7
Therefore,
AB = x + 8 = 7 + 8 = 15 cm
AC = x + 6 = 7 + 6 = 13 cm
Q17. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Solution:
Let a quadrilateral circumscribe a circle with centre O, touching sides at points P, Q, R, S.
Each side is a tangent, so radii to the points of contact are perpendicular to respective sides.
Angles at the centre subtended by two opposite sides correspond to angles formed by pairs of tangents.
Using the property: angle between tangents is supplementary to angle at centre,
we get that the angles subtended by opposite sides at centre add up to 180°.
Hence, opposite sides subtend supplementary angles at the centre. Proved.