## CBSE Class 10 Maths Chapter-10 Circles – Free PDF Download

Free PDF download of Important Questions with Answers for CBSE Class 10 Maths Chapter 10 – Circles prepared by expert Maths teachers from latest edition of CBSE(NCERT) books only by CoolGyan to score more marks in CBSE board examination.

CBSE Class 10 Maths Chapter-10 Circles Important Questions

**CBSE Class 10 Maths Important Questions Chapter 10 – Circles**

**2 Mark Questions**

**1. 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 _______________.**

**Ans. (i)** A tangent to a circle intersects it in __one__ point.

**(ii)** A line intersecting a circle in two points is called a __secan__t.

**(iii)** A circle can have __two__ parallel tangents at the most.

**(iv)** The common point of a tangent to a circle and the circle is called __point of contact__.

**2. 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) cm**

**Ans. (D) ** PQ is the tangent and OP is the radius through the point of contact.

OPQ = [The tangent at any point of a circle is to the radius through the point of contact]

In right triangle OPQ,

OQ^{2} = OP^{2} + PQ^{2} [By Pythagoras theorem]

+ PQ^{2}

144 = 25 + PQ^{2}

PQ^{2} = 144 – 25 = 119

PQ = cm

**3. 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.**

**Ans.**

**4. 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**

**Ans. (A)** OPQ =

[The tangent at any point of a circle is to the radius through the point of contact]

In right triangle OPQ,

OQ^{2} = OP^{2} + PQ^{2} [By Pythagoras theorem]

625 = OP^{2} + 576

OP^{2} = 625 – 576 = 49

OP = 7 cm

**5. In figure, if TP and TQ are the two tangents to a circle with centre O so that POQ = then PTQ is equal to: (A) **

**(B)**

**(C)**

**(D)**

**Ans.**(B) POQ = , OPT = and OQT =

[The tangent at any point of a circle is to the radius through the point of contact]

In quadrilateral OPTQ,

POQ + OPT + OQT + PTQ =

[Angle sum property of quadrilateral]

+ PTQ =

+ PTQ =

PTQ =

**6. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of , then POA is equal to:**

**(A) **

**(B) **

**(C) **

**(D) **

**Ans. (A)** OPQ =

[The tangent at any point of a circle is to the radius through the point of contact]

OPA = BPA[Centre lies on the bisector of the angle between the two tangents]

In OPA,

OAP + OPA + POA = [Angle sum property of a triangle]

+ POA =

+ POA =

POA =

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

**Ans. Given**: PQ is a diameter of a circle with centre O.

The lines AB and CD are the tangents at P and Q respectively.

**To Prove**: AB CD

**Proof**: Since AB is a tangent to the circle at P and OP is the radius through the point of contact.

OPA = ……….(i)

[The tangent at any point of a circle is to the radius through the point of contact]

CD is a tangent to the circle at Q and OQ is the radius through the point of contact.

OQD = ……….(ii)

[The tangent at any point of a circle is to the radius through the point of contact]

From eq. (i) and (ii), OPA = OQD

But these form a pair of equal alternate angles also,

AB CD

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

**Ans.** We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact and the radius essentially passes through the centre of the circle, therefore the perpendicular at the point of contact to the tangent to a circle passes through the centre.

**9. 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.**

**Ans. **We know that the tangent at any point of a circle is to the radius through the point of contact.

OPA =

OA^{2} = OP^{2} + AP^{2} [By Pythagoras theorem]

OP^{2} = 9

OP = 3 cm

**10. 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.**

**Ans.** Let O be the common centre of the two concentric circles.

Let AB be a chord of the larger circle which touches the smaller circle at P.

Join OP and OA.

Then, OPA =

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

OA^{2} = OP^{2} + AP^{2} [By Pythagoras theorem]

+ AP^{2}

25 = 9 + AP^{2}

AP^{2} = 16

AP = 4 cm

Since the perpendicular from the centre of a circle to a chord bisects the chord, therefore

AP = BP = 4 cm

AB = AP + BP = AP + AP = 2AP = 2 x 4 = 8 cm

**11. A quadrilateral ABCD is drawn to circumscribe a circle (see figure). Prove that:**

**AB + CD = AD + BC**

**Ans.** We know that the tangents from an external point to a circle are equal.

AP = AS ……….(i)

BP = BQ ……….(ii)

CR = CQ ……….(iii)

DR = DS ……….(iv)

On adding eq. (i), (ii), (iii) and (iv), we get

(AP + BP) + (CR + DR) = (AS + BQ) + (CQ + DS)

AB + CD = (AS + DS) + (BQ + CQ)

AB + CD = AD + BC

**12. In two concentric circles prove that all chords of the outer circle which touch the inner circle are of equal length.**

**Ans.** AB and CD are two chords of the circle which touch the inner circle at M and N.

Respectively

[AB and CD are two chords of larger circle]

**13. PA and PB are tangents from P to the circle with centre O. At the point M, a tangent is drawn cutting PA at K and PB at N. Prove that KN=AK+BN.**

**Ans. **We know that the lengths of the tangents drawn form an external point to a circle are equal.

(ii) + (iii)

**14. In the given figure, O is the centre of the circle with radius 5 cm and AB||CD. If AB = 6 cm, find OP. Ans. **

OP bisects AB

From right

**15. Prove that the tangents at the end of a chord of a circle make equal angles with the chord. Ans.** In and

BD = DC

And

AD = AD [Common]

[SAS]

[By CPCT]

**16. Find the locus of the centre of circles which touch a given line at a given point.**

**Ans.** Let APB be the given line and let a circle with centre O touch APB at P. Then , let there be another circle with centre O’ which touches the line APB at P.

Thus,

**17. In the given figure, find the perimeter of if AP = 10 cm. Ans.** BC touches the circle at R

Tangents drawn from external point to the circle are equal.

AP = AQ, BR = BP

And CR = CQ

Perimeter of

**18. If PA and PB are tangents drawn from external point P such that PA = 10cm and , find the length of chord AB.**

**Ans.**

[O is centre of circle]

is equilateral triangle

**19. If AB, AC and PQ are tangents in the given figure and AB = 25cm, find the perimeter of Ans. **Perimeter of

**20. Find the locus of the centre of circles which touch a given line at a given point.**

**Ans.** Let APB be the given line and let a circle with centre O touch APB at P. Then , let there be another circle with centre O’ which touches the line APB at P.

Thus,

**21. In the given figure, find the perimeter of if AP = 10 cm. Ans.** BC touches the circle at R

Tangents drawn from external point to the circle are equal.

AP = AQ, BR = BP

And CR = CQ

Perimeter of

**22. If PA and PB are tangents drawn from external point P such that PA = 10 cm and , find the length of chord AB. Ans.**

[O is centre of circle]

is equilateral triangle

**23. If AB, AC and PQ are tangents in the given figure and AB = 25cm, find the perimeter of Ans.** Perimeter of

**24. Find the unknown length x. Ans. **PT is tangent to a circle and PAB is a secant.

**25. In the given figure, OD is perpendicular to the chord AB of a circle whose centre is O. If BC is a diameter, find Ans.** Since BC is a diameter

Also ODAB

[]

[Common]

**26. In the given figure, XP and XQ are tangents from X to the circle with centre O. R is a point on the circle such that ARB is a tangent to the circle prove that XA + AR = XB + BR. Ans. **In the given figure, XP and XQ are tangents from external point

[Length of tangents are equal from external point]

[By (ii) and (iii)]

[By (ii) and (iii)]

**27. Prove that the segment joining the points of contact of two parallel tangents, passes through the centre. Ans.** Given two parallel tangents PQ and RS of a circle with centre O

Draw line OC||RS.

i.e.,

**28. In figure, if OL = 5 cm, OA = 13 cm, then length of AB is Ans.**

=

**29. In the given figure, ABCD is a cyclic quadrilateral and PQ is a tangent to the circle at C. If BD is a diameter, and , find Ans.** BD is a diameter

[Angle in the semi-circle]

**30. Two chords AB and CD of a circle intersect each other at P outside the circle. If AB = 5 cm, BP = 3 cm and PD = 2cm, find CD. Ans.** Two chords AB and CD of a circle intersect each other at P

PA × PB = PC × PD [length of tangent from P]

**31. In the adjoining figure, if AD, AE and BC are tangents to the circle at D, E and F respectively, then prove that 2 AD=AB+BC+CA. Ans.** CD = CF, BE = BF

**32. In figure, PA and PB are tangents from P to the circle with centre O. R is a point on the circle,prove that PC + CR = PD + DR. Ans. **Since length of tangents from an external point to a circle are equal in length

…(i)

And DB = DR

Now PA = PB

PC + CA = PD + DB

PC + CR = PD + DR [By (i)]

**33. The length of tangents from a point A at distance of 26 cm from the centre of the circle is 10cm, what will be the radius of the circle? Ans.** Since tangents to a circle is perpendicular to radius through the point of contact

In right, we have

**34. In the given figure, if TP and TQ are the two tangents to a circle with centre O so that , then find Ans.** Since

**35. In the figure, given below PA and PB are tangents to the circle drawn from an external point P. CD is thethird tangent touching the circle at Q. If PB = 10 cm and CQ = 2 cm, what is the length of PC? Ans.** PA=PB=10 cm

CQ = CA = 2 cm

PC = PA – CA = 10 – 2 = 8 cm