Circles-NCERT Solutions

Class X Math
NCERT Solution for Circle

NCERT TEXTBOOK QUESTIONS SOLVED

EXERCISE 10.1

Q.1. How many tangents can a circle have?

Sol. A circle can have an infinite number of tangents.

Q.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 ........... .

Sol. (i) exactly one (ii) secant (iii) two (iv) point of contact.

Q.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 of PQ is:

(A) 12 cm

(B) 13 cm

cm (D)

Q.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.

Sol. We have the required figure.

Here, l is the given line and a circle with centre 0 is drawn.

The line PT is drawn which is parallel to l and tangent to the circle.

Also, AB is drawn parallel to line l and is a secant to the circle.

EXERCISE 10.2

Q.1. Choose the correct option:

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

(D) 24.5 cm

Sol. QT is a tangent to the circle at T and OT is

radius

Also, OQ = 25 cm and QT = 24 cm

∴ Using Pythagoras theorem, we get

OQ² = QT² + OT²

⇒ OT² = OQ² – QT²

= 25² – 24² = (25 – 24) (25 + 24)

= 1 × 49 = 49 = 7²

⇒ OT = 7

Thus, the required radius is 7 cm.

∴ The correct option is (A).

Q.2. Choose the correct option:

In figure, if TP and TQ are the two tangents to a circle with centre O so that LPOQ = 110°, then LPTQ is equal to

(A) 60°

(B) 70°

(D) 90°

Sol. ∵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).

Q.3. Choose the correct option:

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°

(D) 80°

Sol. Since, O is the centre of the circle and two tangents from P to the circle are PA and PB.

∴ OA ⊥ AP and OB ⊥ BP

⇒ ∠OAP = ∠OBP = 90°

Now, in quadrilateral PAOB, we have:

∠APB + ∠PAO + ∠AOB + ∠PBO = 360°

⇒ 80° + 90° + ∠AOB + 90° = 360°

⇒ 260° + ∠AOB = 360°

⇒ ∠AOB = 360° – 260°

⇒ ∠AOB = 100°

In rt ΔOAP and rt ΔOBP, we have

OP = OP

∠OAF = ∠OBP

OA = OB

∴ Δ OAP ≌ Δ OBP

∴ Their corresponding parts are equal

⇒ ∠POA = ∠POB

Thus, the option (A) is correct.

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

Sol. 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.

PQ ⊥ AB ⇒ ∠APQ = 90°

And PQ ⊥ CD ⇒ ∠PQD = 90°

⇒ ∠APQ = ∠PQD

But they form a pair of alternate angles.

AB ∴∴ CD.

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

Sol. In the figure, the centre of the circle is 0 and tangent AR touches the circle at P. If possible, let PQ be perpendicular to AB such that it is not passing through O.

Join OP.

Since tangent at a point to a circle is perpendicular to the radius through that point,

∴ AB ⊥ OP i.e. ∠OPB = 90° ...(1)

But by construction,

AB ⊥ PQ ⇒ ∠QPB = 90° ...(2)

From (1) and (2),

∠QPB = ∠OPB

which is possible only when O and Q coincide.

Thus, the perpendicular at the point of contact to the tangent passes through the centre.

Q.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.

Sol. ∵The tangent to a circle is perpendicular to the radius through the point of contact.

∠OTA = 90°

Now, in the right ΔOTA, we have:

OP² = OT² + PT²

⇒ 5² = OT² + 4²

⇒ OT² = 5² – 4²

⇒ OT² = (5 – 4) (5 + 4)

⇒ OT² = 1 × 9 = 9 = 3²

⇒ OT = 3

Thus, the radius of the circle is 3 cm.

Q.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.

Sol. In the figure, O is the common centre, of the given concentric circles.

AB is a chord of the bigger circle such that it is a tangent to the smaller circle at P.

Since OP is the radius of the smaller circle through P, the point of contact,

∴ OP ⊥ AB

⇒ ∠APB = 90°

Also, a radius perpendicular to a chord bisects the chord.

Now, in right ΔAPO,

OA² = AP² – OP²

⇒ 5² = AP² – 3²

⇒ AP² = 5² – 3²

⇒ AP² = (5 – 3) (5 + 3) = 2 × 8

⇒ AP² = 16 = (4)²

⇒ AP = 4 cm

Hence, the required length of the chord AB is 8 cm.

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

AB + CD = AD + BC

Sol. Since the sides of quadrilateral ABCD, i.e., AB, BC, CD and DA touch the circle at P, Q, R and S respectively, and the lengths of two tangents to a circle from an external point are equal.

AP = AS

BP = BQ

DR = DS

CR = CQ

Adding them, we get

(AP + BP) + (CR + RD) = (BQ + QC) + (DS + SA)

⇒ AB + CD = BC + DA

which was to be proved.

Q.9. In the figure, XY and X'Y'are two parallel tangents to a circle with centre 0 and another tangent AB with point of contact C intersecting XY at A and XY' at B. Prove that ZAOB = W.

Sol. ∵The tangents drawn to a circle from an external point are equal.

∴ AP = AC

In Δ PAO and Δ AOC, we have:

AO = AO [Common]

OP = OC [Radii of the same circle]

AP = AC

⇒ ΔPAO ≌ ΔAOC

∴ ∠PAO = ∠CAO

∠PAC = 2 ∠CAO ...(1)

Similarly ∠CBQ = 2 ∠CBO ...(2)

Again, we know that sum of internal angles on the same side of a transversal is 180°.

∴ ∠PAC + ∠CBQ = 180°

⇒ 2∠CAO + 2 ∠CBO = 180° [From (1) and (2)]

⇒ 90° + ∠AOB = 180°

⇒ ∠AOB = 180° – 90°

⇒ LAOB = 90°.

Q.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 centre.

Sol. Here, let PA and PB be two tangents drawn from an external point P to a circle with centre O.

Now, in right Δ OAP and right Δ OBP, we have

PA = PB [Tangents to circle from an external point P]

OA = OB [Radii of the same circle]

OP = OP [Comm]

∴ By SSS congruency,

Δ OAP ≌ OBP

∴ Their corresponding parts are equal.

∠OAA = ∠OPB

And ∠AOP = ∠BOP

⇒ ∠APB = 2 ∠OPA and ∠AOS = 2 ∠AOP

But ∠AOP = 90° – LOPA

⇒ 2 ∠AOP = 180° – 2 ∠OPA

⇒ ∠AOB = 180° – ∠APB

⇒ ∠AOB + ∠APB = 180°.

Q.11. Prove that the parallelogram circumscribing a circle is a rhombus. (CBSE 2012, CBSE Delhi 2014)

Sol. We have ABCD, a parallelogram which circumscribes a

circle (i.e., its sides touch the circle) with centre D.

Since tangents to a circle from an external point are equal in length

∴ AP = AS

BP = BQ

CR = CQ

DR = DS

Adding, we get

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

⇒ AB + CD = AD + BC

But AB = CD [opposite sides of ABCD]

and BC = AD

∴ AB + CD = AD + BC ⇒ 2 AB = 2 BC

⇒ AB = BC

Similarly AB = DA and DA = CD

Thus, AB = BC = CD = AD

Hence ABCD is a rhombus.

Q.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 tlic sides AR and AC.

Sol. Here Δ ABC subscribe the circle with centre O.

Also, radius = 4 cm

∵ The sides BC, CA and AB touch the circle at D, E and F respectively.

∴ BF = BD = 8 cm

CE = CD = 6 cm

AF = AE = x cm (say)

⇒ The sides of the triangle are:

14 cm, (x + 6) cm and (x + 8) cm

Perimeter of Δ ABC

= [14 + (x + 6) + (x + 8)] cm

= [14 + 6 + 8 + 2x] cm

= 28 + 2x cm

⇒ Semi perimeter of Δ ABC

∴ S – AB = (14 + x) – (8 + x) = 6

S – BC = (14 + x) – (14) = x

S – AC = (14 + x) – (16 + x) = 8

∴ ar (ΔABC) = ar (Δ OBC) + ar (Δ OCA) + ar (Δ OAB)

= 28 cm² + (2x + 12) cm² + (2x + 16) cm²

= (28 + 12 + 16) + 4x cm²

= (56 + 4x) cm²

From (1) and (2), we have:

Squaring both sides

(14 + x)2 = (14 + x) 3x

⇒ 196 + x² + 28x = 45x + 3x²

⇒ 2x² + 14x – 196 = 0 ⇒ x² + 7x – 98 = 0

⇒ (x – 7) (x + 14) = 0

⇒ Either x – 7 = 0 ⇒ x = 7

or x +14 = 0 ⇒ x = (–14)

But x = (– 14) is not required

∴ x = 7 cm

Thus, AB = 8 + 7 = 15 cm

BC = 8 + 6 = 14 cm

CA = 6 + 7 = 13 cm.

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

Sol. We have a circle with centre O.

A quadrilateral ABCD is such that the sides AB, BC, CD and DA touch the circle at P, Q,,F R and S respectively.

Let us join OP, OQ, OR and OS. We know that two tangents drawn from an external point to a circle subtend equal angles at the centre.

∴ ∠1 = ∠2

∠3 = ∠4

∠5 = ∠6 and ∠7 = ∠8

Also, the sum of all the angles around a point is 360°.

∴ ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360°

∴ 2 [∠1 + ∠8 + ∠5 + ∠4] = 360°

⇒ (∠1 + ∠8 + ∠5 + ∠4) = 180° (1)

And 2 [∠2 + ∠3 + ∠6 + ∠7] = 360°

⇒ (∠2 + ∠3) + (∠6 + ∠7) = 180° (2)

Since, ∠2 + ∠3 = ∠AOB

∠6 + ∠7 = ∠COD

∠1 + ∠8 = ∠AOD

∠4 + ∠5 = ∠BOC

∴ From (1) and (2), we have:

∠AOD + ∠BOC = 180° and

∠AOB + ∠COD = 180°