## 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**

**3 Mark Questions**

**1. Prove that the angel 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. Ans.** OPA = ……….(i)

OCA = ……….(ii)

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

OAPB is quadrilateral.

APB + AOB + OAP + OBP =

[Angle sum property of a quadrilateral]

APB + AOB + + =

[From eq. (i) & (ii)]

APB + AOB =

APB and AOB are supplementary.

**2. Prove that the parallelogram circumscribing a circle is a rhombus.**

**Ans.** **Given**: ABCD is a parallelogram circumscribing a circle.

**To Prove**: ABCD is a rhombus.

**Proof**: Since, 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

AB + AB = AD + AD [Opposite sides of gm are equal]

2AB = 2AD

AB = AD

But AB = CD and AD = BC [Opposite sides of gm]

AB = BC = CD = AD

Parallelogram ABCD is a rhombus.

**3. Two tangents TP and TQ are drawn from an external point T with centre O as shown in figure. If they are inclined to each other at an angle of 100 ^{0}, then what is the value of Ans.** and TQ are tangents and O is the centre of the circle

Quadrilateral OPTQ is cyclic.

**4. 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.** PQ is the chord of the larger circle which touches the smaller circle at the point L. Since PQ is tangent at the point L to the smaller circle with centre O.

is a chord of the bigger circle and

bisects PQ

In

=

Chord PQ = 2PL =8 cm

Length of chord PQ = 8 cm

**5. A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB+CD=AD+BC. Ans.** AP, AS are tangents from a point A (Outside the circle) to the circle.

Similarly, BP = BQ

CQ = CR

DR = DS

Now AB + CD = AP + PB + CR + RD

= AS +BQ + CQ +DS

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

= AD + BC

AB + CD = AD + BC

**6. PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at point T. Find the length TP. Ans.** Join OT.

TP = PQ [tangents from T upon the circle]

And OT bisects PQ

Now

Now

[By AA similarity]

**7. A circle is touching the side BC of at P and touching AB and AC produced at Q and R respectively. Prove that AQ = (perimeter of ). Ans.** We know that the two tangents drawn to a circle from an external point are equal.

Perimeter of

**8. If PA and PB are two tangents drawn from a point P to a circle with centreO touching it at A and B. Prove that OP is the perpendicular bisector of AB. Ans.** Let OP intersect AB at a point C, we have to prove that AC = CB and

are two tangents from a point P to the circle with centre O

[O lies on the bisector of]

In two , ACP and BCP, we have

AP = BP [tangents from P to the circle are equal]

PC = PC [Common]

[Proved]

[By SAS rule]

[CPCT]

And

But

Hence, OP is perpendicular bisector of AB.

**9. In the given figure, PQ is tangent at point R of the circle with centre O. If , find Ans.** Given PQ is tangent at point R and

[Tangent of a circle is perpendicular to Radius]

**10. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle. Ans.** Let the circle touch the sides AB, BC, CD and DA at the points P, Q, R, and S respectively.

Join OP, OQ, OR and OS.

Join OA, OB, OC and OD.

Since the two tangents drawn from an external point subtend equal angles at the centre.

But

[Sum of all angles around a point = 360]

And

And

and

**11. Prove that parallelogram circumscribing a circle is a rhombus. Ans.** Given ABCD is a parallelogram in which all the sides touch a given circle

To prove: ABCD is a rhombus

Proof:

ABCD is a parallelogram

AB=DC and AD = BC

Again AP, AQ are tangents to the circle from the point A

Similarly,

Hence, parallelogram ABCD is a rhombus.

**12. If two tangents are drawn to a circle from an external point then**

**(i) they subtend equal angles at the centre.**

**(ii) they are equally inclined to the segment joining the centre to that point. Ans.** Given on a circle C (O,r), two tangents AP and AQ are drawn from an external point A.

To prove:

**(i)**

**(ii)**

Construction: Join AO, PO and QO

Proof: In and

AP = AQ [Length of the tangents drawn from an external point]

PO = QO [Radii of the same circle]

AO = AO [common]

[By SSS theorem of congruence]

**(i)**[CPCT]

**(ii)**[By CPCT.]

**13. Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that Ans.** Given A circle with centre O and an external point T and two tangents TP and TQ to the circle, where P, Q are the points of contact.

To Prove:

Proof: Let

Since TP, TQ are tangents drawn from point T to the circle.

TP = TQ

TPQ is an isosceles triangle

Since, TP is a tangent to the circle at point of contact P

Thus,

**14. Prove that the lengths of two tangents drawn from an external point to a circle are equal. Ans.** Given: P is an external point to the circle C(O,r).

PQ and PR are two tangents from P to the circle.

To Prove: PQ = PR

Construction: Join OP

Proof:

A tangent to a circle is perpendicular to the radius through the point of contact

Now in right triangles POQ and POR,

OQ = OR [Each radius r]

Hypotenuse. OP = Hypotenuse. OP [common]

[By RHS rule]

**15. The circle of touches the sides BC, CA and AB at D,E and F respectively. If AB = AC, prove that BD = CD. Ans.** Tangents drawn from an external point to a circle are equal in length

AF = AE [Tangents from A] …(i)

BF = BD [Tangents from B] …(ii)

CD = CE [Tangents from C] …(iii)

Adding (i), (ii)and (iii), we get

AF + BF + CD = AE + BD + CE

But AB = AC (given)

CD = BD

**16. A circle touches the side BC of a at a point P and touches AB and AC when produced at Q and R respectively, show that AQ = [Perimeter of ]. Ans.** Since the tangents from an external point to a circle are equal in length,

BP = BQ …(i) [from point B]

CP = CR …(ii) [from point C]

And AQ = AR …(iii) [From point A]

From (iii), we have

AQ = AR

[Using (i) and (ii)]

Now perimeter of

AB + BC +AC = AB +(BP + PC) + AC

= (AB + BP) + (AC +PC)

= 2 (AB + BP) [using (iv)]

=2 (AB + BQ) [using (i)]

= 2 AQ

ÞAQ = (perimeter of )

**17. Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that **

**Ans.** Given: A circle with centre O and an external point T and two tangents TP and TQ to the circle, where P and Q are the points of contact.

To prove:

Proof: Let

In we have

TP = TQ

[Length of the tangents drawn from an external point to a circle are equal]

So, TPQ is an isosceles triangle.

In we have

[Sum of three angles of a is ]

But

[Angle between the tangent and radius of a circle is 90° ]

Now

**18. Prove that the parallelogram circumscribing a circle is a rhombus Ans.** Given: ABCD be the parallelogram circumscribing a circle with centre O such that the sides AB, BC, CD and DA touch a circle at P, Q, R and S respectively.

To prove: ||gm ABCD is a rhombus.

Proof: AP = AS …(i)

BP = BQ …(ii)

CR = CQ …(iii)

DR = DS …(iv)

[Tangents drawn from an external point to a circle are equal]

Adding (i), (ii), (iii) and (iv), we get

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

**19. Prove that the tangents drawn at the ends of a chord of a circle make equal angles with chord. Ans. **Let NM be chord of circle with centre C.

Let tangents at M.N meet at the point O.

Since OM is a tangent

is a tangent

Again in

Thus, tangents make equal angle with the chord.

**20. In the given figure, if AB = AC, prove that BE = EC. Ans.** Since tangents from an exterior point A to a circle are equal in length

Similarly, tangents from an exterior point B to a circle are equal in length

Similarly, for C

CE = CF ………….(3)

Now AB = AC

**21. Find the locus of centre of circle with two intersecting lines. Ans.** Let be two intersection lines.

Let a circle with centre P touch the two lines and at M and N respectively.

PM = PN [Radii of same circle]

P is equidistance from the lines and

Similarly, centre of any other circle which touch the two intersecting lines will be equidistant from and

P lies on a bisector of the angle between and

[The locus of points equidistant from two intersecting lines is the pair of bisectors of the angle between the lines]

Hence, locus of centre of circles which touch two intersecting lines is the pair of bisectors of the angles between the two lines.

**22. In the given figure, a circle is inscribed in a quadrilateral ABCD in which . If AD = 23 cm, AB = 29 cm and DS = 5 cm, find the radius of the circle.**

**Ans. **In the given figure,and OQ ^ BA

Also, OP = OQ =r

is a square

BP = BQ =r

But DR = DS = 5 cm …(i)