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Mathematics · Circles · NCERT Exercises

Exercise 10.2

Complete, independently verified solutions for NCERT Exercise 10.2.

Mathematics · Chapter 10 · NCERT Exercises

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Verified NCERT questions and solutionsSolutions are checked independently and follow the supplied NCERT chapter edition.

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Question

Question 1

multiple choice · radius

From a point Q, a tangent to a circle has length 24 cm and the distance from Q to the centre is 25 cm. Find the radius. Options: (A) 7 cm, (B) 12 cm, (C) 15 cm, (D) 24.5 cm.

Solution

Board-exam working:

Let QT = 24 cm be the tangent and OQ = 25 cm. Radius OT is perpendicular to QT.

OT2=OQ2−QT2OT^2=OQ^2-QT^2
=252−242=25^2-24^2
=625−576=49=625-576=49
OT=7 cmOT=7\text{ cm}

Final answer

Option (A): 7 cm.

Question

Question 2

multiple choice · angle between tangents

In Figure 10.11, TP and TQ are tangents to a circle with centre O and ∠POQ = 110°. Find ∠PTQ. Options: (A) 60°, (B) 70°, (C) 80°, (D) 90°.

Tangents TP and TQ with central angle POQ equal to 110 degrees.
Figure 10-11 from the supplied NCERT chapter.

Solution

Board-exam working:

Radii OP and OQ are perpendicular to tangents TP and TQ.

∠OPT=∠OQT=90∘\angle OPT=\angle OQT=90^\circ

In quadrilateral OPTQ:

∠PTQ+∠POQ+90∘+90∘=360∘\angle PTQ+\angle POQ+90^\circ+90^\circ=360^\circ
∠PTQ+110∘=180∘\angle PTQ+110^\circ=180^\circ
∠PTQ=70∘\angle PTQ=70^\circ

Final answer

Option (B): 70°.

Question

Question 3

multiple choice · angle bisector

Tangents PA and PB from a point P to a circle with centre O are inclined at 80°. Find ∠POA. Options: (A) 50°, (B) 60°, (C) 70°, (D) 80°.

Solution

Board-exam working:

Radii OA and OB are perpendicular to the tangents.

∠OAP=∠OBP=90∘\angle OAP=\angle OBP=90^\circ

In quadrilateral OAPB:

∠AOB+80∘+90∘+90∘=360∘\angle AOB+80^\circ+90^\circ+90^\circ=360^\circ
∠AOB=100∘\angle AOB=100^\circ

Triangles OAP and OBP are congruent, so OP bisects ∠AOB.

∠POA=100∘2=50∘\angle POA=\frac{100^\circ}{2}=50^\circ

Final answer

Option (A): 50°.

Question

Question 4

diameter · parallel tangents

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

Solution

Board-exam working:

Let AB be a diameter. Let lines ℓ and m be the tangents at A and B.

A tangent is perpendicular to the radius through its point of contact.

ℓ⊥OA\ell\perp OA
m⊥OBm\perp OB

OA and OB lie on the same straight line AB.

Two lines perpendicular to the same line are parallel.

∴ ℓ∥m\therefore\ \ell\parallel m

Final answer

Hence, the tangents at the ends of a diameter are parallel.

Question

Question 5

point of contact · radius

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

Solution

Board-exam working:

Let line ℓ touch the circle at P and let O be the centre.

By the tangent-radius theorem, the radius through the point of contact is perpendicular to the tangent.

OP⊥ℓOP\perp\ell

Through a given point P, only one line can be drawn perpendicular to ℓ.

Therefore, the perpendicular to ℓ at P is the line OP itself.

∴ the perpendicular passes through O\therefore\ \text{the perpendicular passes through }O

Final answer

Hence proved.

Question

Question 6

tangent length · radius

The length of a tangent from point A, which is 5 cm from the centre of a circle, is 4 cm. Find the radius.

Solution

Board-exam working:

Let AP = 4 cm be the tangent, O the centre and OP = r.

OP⊥APOP\perp AP

In right triangle OPA:

OA2=OP2+AP2OA^2=OP^2+AP^2
52=r2+425^2=r^2+4^2
r2=25−16=9r^2=25-16=9
r=3 cmr=3\text{ cm}

Final answer

The radius is 3 cm.

Question

Question 7

concentric circles · chord length

Two concentric circles have radii 5 cm and 3 cm. Find the length of a chord of the larger circle that touches the smaller circle.

Solution

Board-exam working:

Let chord AB of the larger circle touch the smaller circle at P. Join OP.

OP⊥ABOP\perp AB

The perpendicular from the centre bisects the chord, so AP = PB.

In right triangle OPA, OA = 5 cm and OP = 3 cm.

AP=OA2−OP2AP=\sqrt{OA^2-OP^2}
=52−32=16=4 cm=\sqrt{5^2-3^2}=\sqrt{16}=4\text{ cm}
AB=2AP=2×4=8 cmAB=2AP=2\times4=8\text{ cm}

Final answer

The chord is 8 cm long.

Question

Question 8

tangential quadrilateral · equal tangents

A quadrilateral ABCD circumscribes a circle as shown in Figure 10.12. Prove that AB + CD = AD + BC.

Quadrilateral ABCD circumscribing a circle at contact points P, Q, R and S.
Figure 10-12 from the supplied NCERT chapter.

Solution

Board-exam working:

Let the circle touch AB, BC, CD and DA at P, Q, R and S respectively.

Tangents from the same external point are equal.

AP=ASAP=AS
BP=BQBP=BQ
CQ=CRCQ=CR
DR=DSDR=DS

Now add the lengths of opposite sides:

AB+CD=(AP+PB)+(CR+RD)AB+CD=(AP+PB)+(CR+RD)
=AS+BQ+CQ+DS=AS+BQ+CQ+DS
=(AS+SD)+(BQ+QC)=(AS+SD)+(BQ+QC)
=AD+BC=AD+BC

Final answer

Hence, AB + CD = AD + BC.

Question

Question 9

parallel tangents · angle bisectors

In Figure 10.13, XY and X′Y′ are parallel tangents to a circle with centre O. Another tangent AB touches at C and meets XY at A and X′Y′ at B. Prove that ∠AOB = 90°.

Parallel tangents XY and X-prime Y-prime and tangent AB touching a circle at C.
Figure 10-13 from the supplied NCERT chapter.

Solution

Board-exam working:

Let P, C and Q be the points of contact of the tangents through A and B.

From A, AP and AC are equal tangents. Therefore, AO bisects ∠PAC.

∠OAB=12∠PAC\angle OAB=\frac12\angle PAC

From B, BQ and BC are equal tangents. Therefore, BO bisects ∠CBQ.

∠ABO=12∠CBQ\angle ABO=\frac12\angle CBQ

Since XY is parallel to X′Y′ and AB is a transversal:

∠PAC+∠CBQ=180∘\angle PAC+\angle CBQ=180^\circ
∠OAB+∠ABO=12(180∘)=90∘\angle OAB+\angle ABO=\frac12(180^\circ)=90^\circ

In triangle AOB:

∠AOB=180∘−90∘=90∘\angle AOB=180^\circ-90^\circ=90^\circ

Final answer

Hence, ∠AOB = 90°.

Question

Question 10

two tangents · supplementary angles

Prove that the angle between two tangents drawn from an external point is supplementary to the angle subtended at the centre by the segment joining the points of contact.

Solution

Board-exam working:

Let PA and PB be tangents from external point P to a circle with centre O.

The radii are perpendicular to the tangents.

∠OAP=∠OBP=90∘\angle OAP=\angle OBP=90^\circ

In quadrilateral OAPB:

∠AOB+∠APB+90∘+90∘=360∘\angle AOB+\angle APB+90^\circ+90^\circ=360^\circ
∠AOB+∠APB=180∘\angle AOB+\angle APB=180^\circ

Thus the angle between the tangents and the central angle subtended by AB are supplementary.

Final answer

Hence proved.

Question

Question 11

circumscribed parallelogram · rhombus

Prove that a parallelogram circumscribing a circle is a rhombus.

Solution

Board-exam working:

Let parallelogram ABCD circumscribe a circle.

For any quadrilateral circumscribing a circle:

AB+CD=AD+BCAB+CD=AD+BC

Opposite sides of a parallelogram are equal.

AB=CDandAD=BCAB=CD\quad\text{and}\quad AD=BC

Substitute these equalities:

AB+AB=AD+ADAB+AB=AD+AD
2AB=2AD2AB=2AD
AB=ADAB=AD

A parallelogram with two adjacent sides equal has all four sides equal.

AB=BC=CD=DAAB=BC=CD=DA

Final answer

Therefore, the parallelogram is a rhombus.

Question

Question 12

circumscribed triangle · Heron's formula

Triangle ABC circumscribes a circle of radius 4 cm. The point of contact D divides BC into BD = 8 cm and DC = 6 cm. Find AB and AC.

Triangle ABC circumscribing a circle of radius 4 centimetres with CD 6 centimetres and DB 8 centimetres.
Figure 10-14 from the supplied NCERT chapter.

Solution

Board-exam working:

Let the circle touch AB and AC at F and E. Tangents from the same point are equal.

BF=BD=8 cmBF=BD=8\text{ cm}
CE=CD=6 cmCE=CD=6\text{ cm}

Let AF = AE = x cm.

AB=x+8,AC=x+6,BC=14AB=x+8,\qquad AC=x+6,\qquad BC=14

The semiperimeter is:

s=(x+8)+(x+6)+142=x+14s=\frac{(x+8)+(x+6)+14}{2}=x+14

Area using the inradius r = 4 cm:

Δ=rs=4(x+14)\Delta=rs=4(x+14)

Area using Heron's formula:

Δ=s(s−AB)(s−AC)(s−BC)\Delta=\sqrt{s(s-AB)(s-AC)(s-BC)}
=(x+14)(6)(8)(x)=\sqrt{(x+14)(6)(8)(x)}

Equating the two expressions and squaring:

16(x+14)2=48x(x+14)16(x+14)^2=48x(x+14)
x+14=3xx+14=3x
x=7x=7
AB=x+8=15 cmAB=x+8=15\text{ cm}
AC=x+6=13 cmAC=x+6=13\text{ cm}

Final answer

AB = 15 cm and AC = 13 cm.

Question

Question 13

tangential quadrilateral · angle bisectors

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

Solution

Board-exam working:

Let quadrilateral ABCD circumscribe a circle with centre O.

The line joining a vertex to O bisects the angle between the two tangents from that vertex.

∠OAB=A2,∠OBA=B2\angle OAB=\frac A2,\quad \angle OBA=\frac B2

In triangle AOB:

∠AOB=180∘−A+B2\angle AOB=180^\circ-\frac{A+B}{2}

Similarly, in triangle COD:

∠COD=180∘−C+D2\angle COD=180^\circ-\frac{C+D}{2}

Since the angle sum of a quadrilateral is 360°:

A+B+C+D=360∘A+B+C+D=360^\circ
∠AOB+∠COD=360∘−A+B+C+D2\angle AOB+\angle COD=360^\circ-\frac{A+B+C+D}{2}
=360∘−180∘=180∘=360^\circ-180^\circ=180^\circ

Similarly:

∠BOC+∠DOA=180∘\angle BOC+\angle DOA=180^\circ

Final answer

Hence, each pair of opposite sides subtends supplementary angles at O.