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 line-segment 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.
