Tangents & Intersecting Chords — Exercise 18(A)

In the given figure, PA, PB and QR are tangents to a circle. If perimeter of the △PQR = 18 cm, the length of tangent PA is :

1. 18 cm

2. 27 cm

3. 9 cm

4. none of these

Answer

Let tangent QR intersect circle at point D.

Given,

Perimeter of the △PQR = 18 cm.

⇒ PQ + QD + RD + PR = 18 .............(1)

We know that,

Two tangents drawn to a circle form an exterior point are equal in length.

∴ PA = PB = x (let), RD = RA and QD = QB.

Substituting above values in equation (1), we get :

⇒ PQ + QB + RA + PR = 18

⇒ PB + PA = 18

⇒ x + x = 18

⇒ 2x = 18

⇒ x = 9 cm.

∴ PA = 9 cm.

Hence, Option 3 is the correct option.

In the given figure, APB is tangent to the inner circle and also a chord of outer circle. Both the circles are concentric. If OA = 10 cm and OP = 6 cm, the length of AB is :

1. 16 cm

2. 10 cm

3. 14 cm

4. 20 cm

Answer

We know that,

Tangent at any point of a circle and the radius through this point are perpendicular to each other.

∴ OP ⊥ AP

In right angle triangle OAP,

By pythagoras theorem,

⇒ OA² = OP² + AP²

⇒ 10² = 6² + AP²

⇒ 100 = 36 + AP²

⇒ AP² = 100 - 36

⇒ AP² = 64

⇒ AP = $\sqrt{64}$ = 8 cm.

Since, AB is the chord to the bigger circle, with center O.

We know that,

Perpendicular from center to the chord, bisects it.

∴ PB = AP = 8 cm.

AB = AP + PB = 8 + 8 = 16 cm.

Hence, Option 1 is the correct option.

A, B and C are three circles which touch each other as shown. Using the information, given in the diagram, we find the length AB as :

1. 6 cm

2. 17 cm

3. (289 - 9 - 2) cm

4. 11 cm

Answer

From figure,

AC = 9 + 6 = 15 cm

BC = 2 + 6 = 8 cm

In right angle triangle ACB,

⇒ AB² = AC² + BC²

⇒ AB² = 15² + 8²

⇒ AB² = 225 + 64

⇒ AB² = 289

⇒ AB = $\sqrt{289}$ = 17 cm.

Hence, Option 2 is the correct option.

BC is a tangent to the circle with center O. OD is radius of the circle. If ∠DOC = 100°, ∠B is equal to :

1. 50°

2. 60°

3. 40°

4. 70°

Answer

From figure,

OD = OA (Radius of same circle)

In △OAD,

∠ODA = ∠OAD = x (let) (As angles opposite to equal sides are equal)

Since, exterior angle in a triangle is equal to the sum of two opposite interior angles.

∴ ∠DOC = ∠ODA + ∠OAD

⇒ 100° = 2x

⇒ x = $\dfrac{100°}{2}$

⇒ x = 50°.

⇒ ∠OAD = 50°.

We know that,

Tangent at any point of a circle and the radius through this point are perpendicular to each other.

∴ ∠BCA = 90°.

In △ABC,

By angle sum property of triangle,

⇒ ∠ABC + ∠BCA + ∠CAB = 180°

⇒ ∠ABC + ∠BCA + ∠OAD = 180° [∵ From figure, ∠CAB = ∠OAD]

⇒ ∠ABC + 90° + 50° = 180°

⇒ ∠ABC + 140° = 180°

⇒ ∠ABC = 180° - 140° = 40°.

Hence, Option 3 is the correct option.

PA and PB are tangents to a circle with center O. If angle BPA = 70°, the angle ACB is :

1. 70°

2. 105°

3. 140°

4. 55°

Answer

Join OA and OB.

We know that,

Tangent at any point of a circle and the radius through this point are perpendicular to each other.

∴ ∠OAP = 90° and ∠OBP = 90°

In quadrilateral OAPB,

⇒ ∠OAP + ∠APB + ∠PBO + ∠BOA = 360°

⇒ 90° + 70° + 90° + ∠BOA = 360°

⇒ ∠BOA + 250° = 360°

⇒ ∠BOA = 360° - 250° = 110°.

We know that,

The angle which an arc of a circle subtends at the center is double that which it subtends at any point on the remaining part of the circumference.

∴ ∠AOB = 2∠ACB

⇒ ∠ACB = $\dfrac{1}{2}$ ∠AOB

⇒ ∠ACB = $\dfrac{1}{2} \times 110°$ = 55°.

Hence, Option 4 is the correct option.

In the given circle with centre O, PA and PB are tangents and ∠OAB = 28°, then ∠APB is :

1. 90°

2. 56°

3. 62°

4. 90° + 28°

Answer

We know that,

Tangent at any point of a circle and the radius through this point are perpendicular to each other.

∴ ∠OAP = 90°

From figure,

⇒ ∠PAB = ∠OAP - ∠OAB = 90° - 28° = 62°

PA and PB are tangents drawn to the circle from the external point P.

∴ PA = PB

We know that,

Angles opposite to equal sides of a triangle are equal.

∴ ∠PAB = ∠PBA = 62°

By angle sum property of triangle PAB,

⇒ ∠PAB + ∠PBA + ∠APB = 180°

⇒ 62° + 62° + ∠APB = 180°

⇒ ∠APB = 180° - 124° = 56°.

Hence, Option 2 is the correct option.

In the above figure, if ∠P = 50°, then reflex angle AOB is :

1. 2 × 50°

2. 180° - 60°

3. 230°

4. none of these

Answer

Join OB.

We know that,

Tangent at any point of a circle and the radius through this point are perpendicular to each other.

∴ ∠OAP = 90° and ∠OBP = 90°

In quadrilateral OAPB,

⇒ ∠OAP + ∠APB + ∠PBO + ∠AOB = 360°

⇒ 90° + 50° + 90° + ∠AOB = 360°

⇒ ∠AOB + 230° = 360°

⇒ ∠AOB = 360° - 230° = 130°.

Reflex ∠AOB = 360° - ∠AOB = 360° - 130° = 230°.

Hence, Option 3 is the correct option.

In the given figure, O is the centre of the circle and AB is a tangent to the circle at B. If AB = 15 cm and AC = 7.5 cm, calculate the radius of the circle.

Answer

Let the radius of the circle to be r cm.

So, AO = AC + OC = (7.5 + r) cm.

In right ∆AOB, we have

⇒ AO² = AB² + OB² [By Pythagoras Theorem]

⇒ (7.5 + r)² = 15² + r²

⇒ 56.25 + r² + 15r = 225 + r²

⇒ 15r = 225 - 56.25

⇒ 15r = 168.75

⇒ r = $\dfrac{168.75}{15}$ = 11.25 cm.

Hence, radius of circle = 11.25 cm.

If the sides of a quadrilateral ABCD touch a circle, prove that AB + CD = BC + AD.

Answer

Let a circle touch the sides AB, BC, CD and DA of quadrilateral ABCD at P, Q, R and S respectively.

We know that,

If two tangents are drawn to a circle from an exterior point, the tangents are equal in length.

As, AP and AS are tangents to the circle from an external point A, we have

AP = AS ......... (1)

Similarly, we also get

BP = BQ ......... (2)

CR = CQ ......... (3)

DR = DS ......... (4)

Adding (1), (2), (3) and (4), we get

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

From figure,

AP + BP = AB, CR + DR = CD, AS + DS = AD and BQ + CQ = BC

⇒ AB + CD = AD + BC

Hence, proved that AB + CD = AD + BC.

From the given figure, prove that :

AP + BQ + CR = BP + CQ + AR.

Also, show that :

AP + BQ + CR = $\dfrac{1}{2}$ x Perimeter of triangle ABC.

Answer

We know that,

If two tangents are drawn to a circle from an exterior point, the tangents are equal in length.

From point B, BQ and BP are the tangents to the circle

BQ = BP ........ (1)

From point A, AP and AR are the tangents to the circle

AP = AR ........ (2)

From point C, CR and CQ are the tangents to the circle

CR = CQ ........ (3)

Adding (1), (2) and (3) we get,

AP + BQ + CR = BP + CQ + AR ......... (4)

Hence, proved that AP + BQ + CR = BP + CQ + AR.

Now, adding AP + BQ + CR to both sides in (4), we get

2(AP + BQ + CR) = AP + BP + CQ + BQ + AR + CR ..........(5)

From figure,

AP + BP = AB, BQ + CQ = BC and AR + CR = AC.

Substituting above value in equation (5), we get :

⇒ 2(AP + BQ + CR) = AB + BC + CA

⇒ AP + BQ + CR = $\dfrac{1}{2}$(AB + BC + CA).

⇒ AP + BQ + CR = $\dfrac{1}{2}$ (Perimeter of △ABC). [∵ Perimeter = AB + BC + CA]

Hence, proved that AP + BQ + CR = $\dfrac{1}{2}$ (Perimeter of △ABC).

In the given figure, if AB = AC then prove that BQ = CQ.

Answer

As, from point A, AP and AR are the tangents to the circle.

We know that,

If two tangents are drawn to a circle from an exterior point, the tangents are equal in length.

So, we have AP = AR ..........(1)

From point B, BP and BQ are the tangents to the circle.

∴ BP = BQ ..........(2)

From point C, CQ and CR are the tangents to the circle.

∴ CQ = CR ............(3)

Adding equations (1), (2) and (3), we get :

⇒ AP + BP + CQ = AR + BQ + CR

⇒ (AP + BP) + CQ = (AR + CR) + BQ

⇒ AB + CQ = AC + BQ

Given,

AB = AC

∴ BQ = CQ.

Hence, proved that BQ = CQ.

Radii of two circles are 6.3 cm and 3.6 cm. State the distance between their centers if :

(i) they touch each other externally,

(ii) they touch each other internally.

Answer

Let O be the center of the circle with radius = 6.3 cm and O' be the center of circle with radius = 3.6 cm.

(i) When the two circles touch each other at P externally. O' and O are the centers of the circles. Join O'P and OP.

So, O'P = 6.3 cm, OP = 3.6 cm

Hence, the distance between their centres (O'O) is given by

O'O = O'P + OP = 6.3 + 3.6 = 9.9 cm.

Hence, distance between their centers if they touch each other externally is 9.9 cm.

(ii) When the two circles touch each other at P internally, O and O' are the centers of the circles. Join OP and O'P.

So, O'P = 6.3 cm, OP = 3.6 cm.

Hence, the distance between their centres (O'O) is given by

O'O = O'P - OP = 6.3 - 3.6 = 2.7 cm.

Hence, distance between their centers if they touch each other internally is 2.7 cm.

In the given Figure, two circles touch each other externally at point P. AB is the direct common tangent of these circles. Prove that :

(i) tangent at point P bisects AB.

(ii) angle APB = 90°.

Answer

(i) We know that,

If two tangents are drawn to a circle from an exterior point, the tangents are equal in length.

From figure,

TA and TP are the tangents to the circle with center O.

∴ TA = TP ...........(1)

TB and TP are the tangents to the circle with center O'.

∴ TB = TP ...........(2)

From (1) and (2) we get :

TA = TB.

Hence, proved that tangent at point P bisects AB.

(ii) In △ATP,

TA = TP [Proved above]

∴ ∠TAP = ∠TPA .........(1) [∵ angles opposite to equal sides are equal.]

In △BTP,

TB = TP [Proved above]

∴ ∠TBP = ∠TPB ..........(2) [∵ angles opposite to equal sides are equal.]

Adding (1) and (2), we get :

∠TAP + ∠TBP = ∠TPA + ∠TPB

∠TAP + ∠TBP = ∠APB ...........(3)

In △ABP,

⇒ ∠APB + ∠BAP + ∠ABP = 180° [Angle sum property of triangle]

⇒ ∠APB + ∠TAP + ∠TBP = 180° [From figure, ∠TAP = ∠BAP and ∠TBP = ∠ABP.]

⇒ ∠APB + ∠APB = 180°

⇒ 2∠APB = 180°

⇒ ∠APB = 90°.

Hence, proved that ∠APB = 90°.

Tangents AP and AQ are drawn to a circle, with center O, from an exterior point A. Prove that :

∠PAQ = 2∠OPQ

Answer

We know that,

The tangent at any point of a circle and the radius through this point are perpendicular to each other.

In quadrilateral OPAQ,

∠OPA = ∠OQA = 90°.

∠OPA + ∠OQA + ∠POQ + ∠PAQ = 360° [∵ Sum of angles in quadrilateral = 360°]

90° + 90° + ∠POQ + ∠PAQ = 360°

∠POQ + ∠PAQ = 360° - 180°

∠POQ + ∠PAQ = 180° ..........(1)

In △OPQ,

OP = OQ [Radius of same circle]

∴ ∠OPQ = ∠OQP [Angles opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠OPQ + ∠OQP + ∠POQ = 180°

⇒ ∠OPQ + ∠OPQ + ∠POQ = 180°

⇒ 2∠OPQ + ∠POQ = 180° .........(2)

From (1) and (2) we get,

∠POQ + ∠PAQ = 2∠OPQ + ∠POQ

⇒ ∠PAQ = 2∠OPQ.

Hence, proved that ∠PAQ = 2∠OPQ.

ABC is a right angled triangle with AB = 12 cm and AC = 13 cm. A circle, with center O, has been inscribed inside the triangle. Calculate the value of x, the radius of the inscribed circle.

Answer

Let AB touches the circle at L, AC at N and BC at M.

From figure,

LBMO is a square.

LB = BM = OM = OL = x.

AL = AB - LB = (12 - x) cm.

AL = AN = (12 - x) cm. [∵ Tangents from exterior point are equal in length.]

Since, ABC is a right angled triangle,

∴ AC² = AB² + BC² [By pythagoras theorem]

⇒ 13² = 12² + BC²

⇒ BC² = 13² - 12²

⇒ BC² = 169 - 144

⇒ BC² = 25

⇒ BC = $\sqrt{25}$

⇒ BC = 5 cm.

From figure,

MC = BC - BM = (5 - x) cm.

Also,

CN = CM = (5 - x) cm. [∵ Tangents from exterior point are equal in length.]

Also,

⇒ AC = AN + CN

⇒ 13 = (12 - x) + (5 - x)

⇒ 13 = 17 - 2x

⇒ 2x = 17 - 13

⇒ 2x = 4

⇒ x = $\dfrac{4}{2}$

⇒ x = 2 cm.

Hence, x = 2.

In the given figure, PT touches the circle with center O at point R. Diameter SQ is produced to meet the tangent TR at P.

Given ∠SPR = x° and ∠QRP = y°;

prove that :

(i) ∠ORS = y°

(ii) write an expression connecting x and y.

Answer

(i) From figure,

⇒ ∠QRP = ∠OSR = y° [Angles in alternate segment are equal]

⇒ OS = OR (Radius of same circle)

As, angles opposite to equal sides are equal,

∴ ∠ORS = ∠OSR = y°.

Hence, proved that ∠ORS = y°.

(ii) From figure,

∠ORP = 90° [As, tangent to a point and radius from that point are perpendicular to each other.]

⇒ ∠ORQ = ∠ORP - ∠QRP = 90° - y° ...........(1)

OQ = OR (Radius of same circle)

As, angles opposite to equal sides are equal,

∴ ∠OQR = ∠ORQ = 90° - y°

In △PQR,

⇒ ∠OQR = ∠QPR + ∠QRP (As exterior angle in a trinagle is equal to the sum of two opposite interior angles.)

⇒ 90° - y° = x° + y°

⇒ x° + 2y° = 90°.

Hence, x + 2y = 90°.

PT is a tangent to the circle at T. If ∠ABC = 70° and ∠ACB = 50°; calculate :

(i) ∠CBT

(ii) ∠BAT

(iii) ∠APT

Answer

Join AT and BT.

(i) TC is the diameter of the circle.

Since, angle in a semi-circle is a right angle.

∴ ∠CBT = 90°.

Hence, ∠CBT = 90°.

(ii) In cyclic quadrilateral ATBC,

⇒ ∠CBT + ∠CAT = 180° (∵ Sum of opposite angles of a cyclic quadrilateral = 180°)

⇒ 90° + ∠CAT = 180°

⇒ ∠CAT = 180° - 90°

⇒ ∠CAT = 90°.

In △ABC,

⇒ ∠CBA + ∠CAB + ∠ACB = 180° [By angle sum property of triangle]

⇒ 70° + ∠CAB + 50° = 180°

⇒ ∠CAB + 120° = 180°

⇒ ∠CAB = 180° - 120°

⇒ ∠CAB = 60°.

From figure,

∠BAT = ∠CAT - ∠CAB = 90° - 60° = 30°.

Hence, ∠BAT = 30°.

(iii) From figure,

∠BTX = ∠BAT = 30° [Angle in same segment are equal]

∠PBT = ∠CBT - ∠CBA = 90° - 70° = 20°.

⇒ ∠PTB = 180° - ∠BTX = 180° - 30° = 150°.

In △PBT,

⇒ ∠PBT + ∠PTB + ∠APT = 180° [By angle sum property of triangle]

⇒ 20° + 150° + ∠APT = 180°

⇒ ∠APT + 170° = 180°

⇒ ∠APT = 180° - 170°

⇒ ∠APT = 10°.

Hence, ∠APT = 10°.

In the given figure, O is the center of the circumcircle ABC. Tangents A and C intersect at P. Given angle AOB = 140° and angle APC = 80°; find the angle BAC.

Answer

Join OC

∴ PA and PC are the tangents

∴ OA ⊥ PA and OC ⊥ PC

In quadrilateral APCO,

⇒ ∠APC + ∠AOC = 180°

⇒ 80° + ∠AOC = 180°

⇒ ∠AOC = 180° - 80°

⇒ ∠AOC = 100°

From figure,

∠BOC = 360° - (∠AOB + ∠AOC)

= 360° - (140° + 100°)

= 360° - 240° = 120°.

We know that,

The angle at the centre of a circle is twice the angle at the circumference, subtended by the same arc.

Now arc BC subtends ∠BOC at the centre and ∠BAC at the remaining part of the circle.

∴ ∠BAC = $\dfrac{1}{2}$∠BOC = $\dfrac{1}{2} \times 120°$ = 60°.

Hence, ∠BAC = 60°.

In the given figure, PQ is a tangent to the circle at A. AB and AD are bisectors of ∠CAQ and ∠PAC. If ∠BAQ = 30°, prove that : BD is diameter of the circle.

Answer

From figure,

∠CAB = ∠BAQ = 30° (AB is angle bisector of ∠CAQ)

⇒ ∠CAQ = 2∠BAQ = 60°.

From figure,

⇒ ∠CAQ + ∠PAC = 180° [Linear pair]

⇒ 60° + ∠PAC = 180°

⇒ ∠PAC = 180° - 60°

⇒ ∠PAC = 120°.

⇒ ∠PAC = 2∠CAD (AD is angle bisector of ∠PAC)

⇒ 120° = 2∠CAD

⇒ ∠CAD = $\dfrac{120°}{2}$

⇒ ∠CAD = 60°.

From figure,

∠DAB = ∠CAD + ∠CAB = 60° + 30° = 90°.

Thus BD, subtends 90° on the circle. Since, angle in semi-circle is a right angle.

Hence, BD is the diameter of the circle.