NCERT Solution for Class 10 Mathematics Chapter 10 Circles
Chapter Name  NCERT Solution for Class 10 Maths Chapter 10 Circles 
Topics Covered 

Related Study 

Short Revision for Circles
1. Only one tangent is possible at a point on the circle.
2. No tangent is possible from a point in the interior of a circle, to the circle.
3. At most two tangents can be drawn from a point in the exterior of a circle, to the circle.
4. The lengths of the two tangents drawn from an external point to a circle are equal.
5. Tangent at any point on a circle is perpendicular to the radius passing through the point of contact.
NCERT Exercise Solutions
Exercise: 10.1
1. How many tangents can a circle have?
Solution
A circle can have infinitely many tangents.
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 …………
Solution
(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) √119 cm
Solution
(D)
4. 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.
Exercise: 10.2
In Q.1 to 3, choose the correct option and give justification.
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
2. In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to
(A) 60°
(B) 70°
(C) 80°
(D) 90°
Solution
∵ TQ and TP are tangents to a circle with centre O.
such that ∠POQ = 110°
∴ OP⊥ PT and OQ ⊥ QT
⇒ ∠OPT = 90 and ∠OQT = 90°
Now, in the quadrilateral TPOQ, we get
∴ ∠PTQ + 90° + 110° + 90° = 360°
⇒ ∠PTQ + 290° = 360°
⇒ ∠PTQ = 360°  290° = 70°
Thus, the correct option is (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 ∠ POA is equal to
(A) 50°
(B) 60°
(C) 70°
(D) 80°
4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Solution
In the figure, we have :
PQ is diameter of the given circle and O is its centre.
Let tangents AB and CD be drawn at the end points of the diameter PQ.
Since the tangent at a point to a circle is perpendicular to the radius through the point.
But they form a pair of alternate angles.
∴ AB ॥ CD.
5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the center.
Solution
Let perpendicular at the point of contact to the tangent does not pass through at centre.
O'P⊥PT .....(i) [Given]
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.
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.
8. A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that AB + CD = AD + BC
Consider AB + CD = (AP + PB) + (CR + RD)
= AS + BQ + CQ + DS
= (AS + DS)+ (BQ + CQ)
= AD + BC
9. 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 ∠ AOB = 90°.
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 linesegment joining the points of contact at the center.
11. Prove that the parallelogram circumscribing a circle is a rhombus.
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 (see figure). Find the sides AB and AC .
Solution
Let the circumcircle touches AB and AC at E and F respectively.
Join OA, OB, OC, OE and OF.
13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.