## NCERT Solutions for Class 10 Maths Exercise 10.2 Chapter 10 Circles â€“ FREE PDF Download

NCERT Class 10 Maths Ch 10 is one of the most important ones in the NCERT syllabus. Duly following NCERT Solutions for Class 10 Maths

Chapter 10 ensures you that there will be no hindrance when you opt for more advanced branches of Maths. This is where CoolGyan comes in. Our free Class 10 Circles solutions will help you understand the chapter thoroughly.

# NCERT Solutions for Class 10 Maths Chapter 10 â€“ Circles

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 Â (B) 12 cm Â (C) 15 cm Â (D) 24.5 cm

Â OPQ =Â

[The tangent at any point of a circle isÂ Â to the radius

through the point of contact]

In right triangle OPQ,

[By Pythagoras theorem]

Â

Â = 625 â€“ 576 = 49

OP = 7 cm

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 Â (B) 12 cm Â (C) 15 cm Â (D) 24.5 cm

**Ans.**Â (A)

Â OPQ =Â

[The tangent at any point of a circle isÂ Â to the radius

through the point of contact]

In right triangle OPQ,

[By Pythagoras theorem]

Â

Â = 625 â€“ 576 = 49

OP = 7 cm

2. In figure, if TP and TQ are the two tangents to a circle with centre O so thatÂ
POQ =Â
Â thenÂ
PTQ is equal to:

(A)Â Â (B)Â Â (C)Â Â (D)Â

POQ =Â ,Â OPT =Â Â andÂ OQT =Â

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

In quadrilateral OPTQ,

POQ +Â OPT +Â OQT +Â PTQ =Â

[Angle sum property of quadrilateral]

Â +Â PTQ =Â

Â Â +Â PTQ =Â

â‡’â‡’Â Â Â âˆ âˆ PTQ = 360

Â PTQ =Â

(A)Â Â (B)Â Â (C)Â Â (D)Â

**Ans.**Â (B)

POQ =Â ,Â OPT =Â Â andÂ OQT =Â

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

In quadrilateral OPTQ,

POQ +Â OPT +Â OQT +Â PTQ =Â

[Angle sum property of quadrilateral]

Â +Â PTQ =Â

Â Â +Â PTQ =Â

â‡’â‡’Â Â Â âˆ âˆ PTQ = 360

^{o}Â â€“ 290

^{o}

Â PTQ =Â

3. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle ofÂ
, thenÂ
POA is equal to:

(A)Â Â (B)Â Â (C)Â Â (D)Â

Â OAP =Â

[The tangent at any point of a circle isÂ Â to the radius

through the point of contact]

OPA =12âˆ 12âˆ Â BPA =Â 12Ã—80âˆ˜=40âˆ˜12Ã—80âˆ˜=40âˆ˜

[Centre lies on the bisector of the

angle between the two tangents]

InÂ OPA,

OAP +Â OPA +Â POA =Â

[Angle sum property of a triangle]

Â +Â POA =Â

Â Â +Â POA =Â

Â POA =Â

(A)Â Â (B)Â Â (C)Â Â (D)Â

**Ans.**Â (A)

Â OAP =Â

[The tangent at any point of a circle isÂ Â to the radius

through the point of contact]

OPA =12âˆ 12âˆ Â BPA =Â 12Ã—80âˆ˜=40âˆ˜12Ã—80âˆ˜=40âˆ˜

[Centre lies on the bisector of the

angle between the two tangents]

InÂ OPA,

OAP +Â OPA +Â POA =Â

[Angle sum property of a triangle]

Â +Â POA =Â

Â Â +Â POA =Â

Â POA =Â

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

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

To Prove: ABÂ Â CD

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

OPA =Â â€¦â€¦â€¦.(i)

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

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

OQD =Â â€¦â€¦â€¦.(ii)

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

From eq. (i) and (ii),Â OPA =Â OQD

But these form a pair of equal alternate angles also,

Â AB ||Â CD

**Ans.**Â 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: ABÂ Â CD

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

OPA =Â â€¦â€¦â€¦.(i)

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

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

OQD =Â â€¦â€¦â€¦.(ii)

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

From eq. (i) and (ii),Â OPA =Â OQD

But these form a pair of equal alternate angles also,

Â AB ||Â CD

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

If possible, let PQ be perpendicular to AB, not passing through O.

Join OP.

Since tangnet at a point to a circle is perpendicular to the radius through the point.

Therefore, ABÂ âŠ¥âŠ¥Â OPÂ Â Â Â Â Â Â Â Â Â â‡’â‡’Â Â Â Â Â Â Â Â Â Â âˆ âˆ OPB = 90

Also,Â âˆ âˆ QPB = 90

Therefore,Â Â âˆ âˆ QPB =Â âˆ âˆ OPB, which is not possible as a part cannot be equal to whole.

Thus, it contradicts our supposition.

Hence, the perpendicular at the point of contact to the tangent to a circle passes through the centre.

**Ans.**Â Let AB be the tangent drawn at the point P on the circle with O.

If possible, let PQ be perpendicular to AB, not passing through O.

Join OP.

Since tangnet at a point to a circle is perpendicular to the radius through the point.

Therefore, ABÂ âŠ¥âŠ¥Â OPÂ Â Â Â Â Â Â Â Â Â â‡’â‡’Â Â Â Â Â Â Â Â Â Â âˆ âˆ OPB = 90

^{o}

Also,Â âˆ âˆ QPB = 90

^{o}Â Â Â Â Â Â Â Â Â Â Â Â [By construction]

Therefore,Â Â âˆ âˆ QPB =Â âˆ âˆ OPB, which is not possible as a part cannot be equal to whole.

Thus, it contradicts our supposition.

Hence, the perpendicular at the point of contact to the tangent to a circle passes through the centre.

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.

Â OPA =Â

Â

[By Pythagoras theorem]

Â

Â = 9

Â OP = 3 cm

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

Â OPA =Â

Â

[By Pythagoras theorem]

Â

Â = 9

Â OP = 3 cm

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.

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

Join OP and OA.

Then,Â Â OPA =Â

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

Â OA

[By Pythagoras theorem]

Â

Â = 16

Â AP = 4 cm

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

AP = BP = 4 cm

Â AB = AP + BP

= AP + AP = 2AP

=Â = 8 cm

**Ans.**Â 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,Â Â OPA =Â

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

Â OA

^{2}Â = OP

^{2}Â + AP

^{2}

[By Pythagoras theorem]

Â

Â = 16

Â AP = 4 cm

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

AP = BP = 4 cm

Â AB = AP + BP

= AP + AP = 2AP

=Â = 8 cm

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

AB + CD = AD + BC

AP = AS â€¦â€¦â€¦.(i)

BP = BQ â€¦â€¦â€¦.(ii)

CR = CQ â€¦â€¦â€¦.(iii)

DR = DSâ€¦â€¦â€¦.(iv)

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

(AP + BP) + (CR + DR)

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

Â AB + CD = (AS + DS) + (BQ + CQ)

Â AB + CD = AD + BC

AB + CD = AD + BC

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

AP = AS â€¦â€¦â€¦.(i)

BP = BQ â€¦â€¦â€¦.(ii)

CR = CQ â€¦â€¦â€¦.(iii)

DR = DSâ€¦â€¦â€¦.(iv)

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

(AP + BP) + (CR + DR)

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

Â AB + CD = (AS + DS) + (BQ + CQ)

Â AB + CD = AD + BC

9. In figure, XY and Xâ€™Yâ€™ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and Xâ€™Yâ€™ at B. Prove thatÂ
AOB =Â

To Prove:Â AOB =Â

Construction: Join OC

Proof:Â OPA =Â â€¦â€¦â€¦.(i)

OCA =Â â€¦â€¦â€¦.(ii)

[Tangent at any point of a circle isÂ Â to

the radius through the point of contact]

In right angled triangles OPA and OCA,

OPA =Â OCA = 90

OA = OA [Common]

AP = AC [Tangents from an external

point to a circle are equal]

OPAÂ OCA

[RHS congruence criterion]

OAP =Â OAC [By C.P.C.T.]

OAC =Â PAB â€¦â€¦â€¦.(iii)

Similarly,Â OBQ =Â OBC

OBC =Â QBA â€¦â€¦â€¦.(iv)

XYÂ Â Xâ€™Yâ€™ and a transversal AB intersects them.

PAB +Â QBA =Â

[Sum of the consecutive interior angles on the same side of the transversal isÂ ]

PAB +Â QBA

=Â â€¦â€¦â€¦.(v)

Â OAC +Â OBC =Â

[From eq. (iii) & (iv)]

InÂ AOB,

OAC +Â OBC +Â AOB =Â

[Angel sum property of a triangle]

Â +Â AOB =Â [From eq. (v)]

Â AOB =Â

Hence proved.

**Ans.**Â Given: In figure, XY and Xâ€™Yâ€™ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and Xâ€™Yâ€™ at B.

To Prove:Â AOB =Â

Construction: Join OC

Proof:Â OPA =Â â€¦â€¦â€¦.(i)

OCA =Â â€¦â€¦â€¦.(ii)

[Tangent at any point of a circle isÂ Â to

the radius through the point of contact]

In right angled triangles OPA and OCA,

OPA =Â OCA = 90

^{o}

OA = OA [Common]

AP = AC [Tangents from an external

point to a circle are equal]

OPAÂ OCA

[RHS congruence criterion]

OAP =Â OAC [By C.P.C.T.]

OAC =Â PAB â€¦â€¦â€¦.(iii)

Similarly,Â OBQ =Â OBC

OBC =Â QBA â€¦â€¦â€¦.(iv)

XYÂ Â Xâ€™Yâ€™ and a transversal AB intersects them.

PAB +Â QBA =Â

[Sum of the consecutive interior angles on the same side of the transversal isÂ ]

PAB +Â QBA

=Â â€¦â€¦â€¦.(v)

Â OAC +Â OBC =Â

[From eq. (iii) & (iv)]

InÂ AOB,

OAC +Â OBC +Â AOB =Â

[Angel sum property of a triangle]

Â +Â AOB =Â [From eq. (v)]

Â AOB =Â

Hence proved.

10. 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.

OBP =Â â€¦â€¦â€¦.(ii)

[Tangent at any point of a circle isÂ Â to

the radius through the point of contact]

Â OAPB is quadrilateral.

Â APB +Â AOB +Â OAP +Â OBP =Â

[Angle sum property of a quadrilateral]

Â APB +Â AOB +Â Â +Â =Â

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

Â APB +Â AOB =Â

Â APB andÂ AOB are supplementary.

**Ans.**Â OAP =Â â€¦â€¦â€¦.(i)

OBP =Â â€¦â€¦â€¦.(ii)

[Tangent at any point of a circle isÂ Â to

the radius through the point of contact]

Â OAPB is quadrilateral.

Â APB +Â AOB +Â OAP +Â OBP =Â

[Angle sum property of a quadrilateral]

Â APB +Â AOB +Â Â +Â =Â

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

Â APB +Â AOB =Â

Â APB andÂ AOB are supplementary.

11. Prove that the parallelogram circumscribing a circle is a rhombus.

To Prove: ABCD is a rhombus.

Proof: Since, the tangents from an external point to a circle are equal.

AP = AS â€¦â€¦â€¦.(i)

BP = BQ â€¦â€¦â€¦.(ii)

CR = CQ â€¦â€¦â€¦.(iii)

DR = DSâ€¦â€¦â€¦.(iv)

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

(AP + BP) + (CR + DR)

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

Â AB + CD = (AS + DS) + (BQ + CQ)

Â AB + CD = AD + BC

Â AB + AB = AD + AD

[Opposite sides ofÂ gm are equal]

Â 2AB = 2AD

Â AB = AD

But AB = CD and AD = BC

[Opposite sides ofÂ gm]

AB = BC = CD = AD

Â Parallelogram ABCD is a rhombus.

**Ans.**Â 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 â€¦â€¦â€¦.(i)

BP = BQ â€¦â€¦â€¦.(ii)

CR = CQ â€¦â€¦â€¦.(iii)

DR = DSâ€¦â€¦â€¦.(iv)

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

(AP + BP) + (CR + DR)

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

Â AB + CD = (AS + DS) + (BQ + CQ)

Â AB + CD = AD + BC

Â AB + AB = AD + AD

[Opposite sides ofÂ gm are equal]

Â 2AB = 2AD

Â AB = AD

But AB = CD and AD = BC

[Opposite sides ofÂ gm]

AB = BC = CD = AD

Â Parallelogram ABCD 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.

Since BD = 8 cm

BE = 8 cm

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

Since CD = 6 cm

CF = 6 cm

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

Let AE = AF =Â

Since OD = OE = OF = 4 cm

[Radii of a circle are equal]

Â Semi-perimeter ofÂ ABC =Â Â =Â Â cm

Â Area ofÂ ABC =Â

=Â

=Â Â cm

Now, Area ofÂ ABC = Area ofÂ OBC + Area ofÂ OCA + Area ofÂ OAB

=Â

=Â

Â =Â

Â =Â

Squaring both sides,

Â

Â

Â

AB =Â Â = 7 + 8 = 15 cm

And AC =Â Â = 7 + 6 = 13 cm

**Ans.**Â Join OE and OF. Also join OA, OB and OC.

Since BD = 8 cm

BE = 8 cm

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

Since CD = 6 cm

CF = 6 cm

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

Let AE = AF =Â

Since OD = OE = OF = 4 cm

[Radii of a circle are equal]

Â Semi-perimeter ofÂ ABC =Â Â =Â Â cm

Â Area ofÂ ABC =Â

=Â

=Â Â cm

^{2}

Now, Area ofÂ ABC = Area ofÂ OBC + Area ofÂ OCA + Area ofÂ OAB

=Â

=Â

Â =Â

Â =Â

Squaring both sides,

Â

Â

Â

AB =Â Â = 7 + 8 = 15 cm

And AC =Â Â = 7 + 6 = 13 cm

13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

To prove: (i)Â AOB +Â COD =Â (ii)Â BOC +Â AOD =Â

Construction: Join OP, OQ, OR and OS.

Proof: Since tangents from an external point to a circle are equal.

AP = AS,

BP = BQ â€¦â€¦â€¦.(i)

CQ = CR

DR = DS

InÂ OBP andÂ OBQ,

OP = OQ [Radii of the same circle]

OB = OB [Common]

BP = BQ [From eq. (i)]

Â OPBÂ OBQ [By SSS congruence criterion]

Â [By C.P.C.T.]

Similarly,Â Â Â

Since, the sum of all the angles round a point is equal toÂ

Â

Â

Â

Â AOB +Â COD =Â

Similarly, we can prove that

BOC +Â AOD =Â

**Ans.**Â Given: ABCD is a quadrilateral circumscribing a circle whose centre is O.

To prove: (i)Â AOB +Â COD =Â (ii)Â BOC +Â AOD =Â

Construction: Join OP, OQ, OR and OS.

Proof: Since tangents from an external point to a circle are equal.

AP = AS,

BP = BQ â€¦â€¦â€¦.(i)

CQ = CR

DR = DS

InÂ OBP andÂ OBQ,

OP = OQ [Radii of the same circle]

OB = OB [Common]

BP = BQ [From eq. (i)]

Â OPBÂ OBQ [By SSS congruence criterion]

Â [By C.P.C.T.]

Similarly,Â Â Â

Since, the sum of all the angles round a point is equal toÂ

Â

Â

Â

Â AOB +Â COD =Â

Similarly, we can prove that

BOC +Â AOD =Â