Tangent Properties of Circles — Multiple Choice Questions

In two concentric circles, a chord of larger circle which is ________ to smaller circle is bisected at the point of contact.

1. secant

2. tangent

3. chord

4. diameter

Answer

In two concentric circles, a chord of larger circle which is tangent to smaller circle is bisected at the point of contact.

Hence, option 2 is the correct option.

The length of the tangent drawn to a circle of radius 8 cm, from a point which is at a distance of 10 cm from the centre of the circle is :

1. 6 cm

2. 7 cm

3. 9 cm

4. 2 cm

Answer

We know that,

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

In right ∆OTP, we have

OT² = PT² + OP²

PT² = OT² - OP²

PT² = 100 - 64

PT² = 36

PT = $\sqrt{36}$ = 6 cm.

Hence, option 1 is the correct option.

There are ________ two tangents to a circle passing through a point lying outside the circle.

1. at least

2. at most

3. exactly

4. maximum

Answer

A standard theorem in circle states:

From a point lying outside a circle, exactly two tangents can be drawn to the circle.

Hence, option 3 is the correct option.

If two tangents are drawn from an external point to a circle, then the tangents are equally inclined to the line joining the point and the centre of the circle, i.e., the centre lies on the ________ of the angle between the two tangents.

1. perpendicular

2. bisector

3. perpendicular bisector

4. none of these

Answer

When two tangents are drawn from an external point to a circle, the line joining the external point and the centre of the circle bisects the angle between the two tangents.

Hence, option 2 is the correct option.

From a point M, the length of the tangent to a circle is 24 cm and the distance of M from the centre is 25 cm. The radius of the circle is :

1. 7 cm

2. 12 cm

3. 24.5 cm

4. 12.5 cm

Answer

We know that,

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

In right ∆MOT, we have

MO² = MT² + OT²

OT² = MO² - MT²

OT² = 25² - 24²

OT² = 625 - 576

OT² = 49

OT = $\sqrt{49}$

OT = 7 cm.

Hence, option 1 is the correct option.

PQ is a tangent to a circle at point P. Centre of the circle is O. If ΔOPQ is an isosceles triangle, then ∠QOP =

1. 30°

2. 60°

3. 45°

4. 90°

Answer

Let QP = OP

We know that,

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

∠OPQ = 90°

∠QOP = ∠OQP = x [Anglers opposite to equal sides are equal in a triangle]

By angles sum triangle property,

x + x + 90° = 180°

2x + 90° = 180°

2x = 180° - 90°

2x = 90°

x = $\dfrac{90°}{2}$

x = 45°

∠QOP = 45°

Hence, option 3 is the correct option.

In the figure XY and XZ are tangents at points Y and Z respectively to a circle with centre O. If C is a point on the circle and ∠ZXY = 40°, then ∠ZCY = ?

1. 80°

2. 70°

3. 140°

4. 40°

Answer

From figure,

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

∠OZX = 90°

∠OYX = 90°

∠ZXY + ∠ZOY + ∠OZX + ∠OYX = 360° [Angle sum property of quadrilateral]

40° + 90° + 90° + ∠ZOY = 360°

220° + ∠ZOY = 360°

∠ZOY = 360° - 220°

∠ZOY = 140°.

We know that,

Angle at the centre is double the angle at a point on the remaining part of the circle.

∠ZCY = $\dfrac{1}{2}$ ∠ZOY

∠ZCY = $\dfrac{140°}{2}$

∠ZCY = 70°.

Hence, option 2 is the correct option.

In the given figure, if sides AB, BC, CD and DA of a quadrilateral ABCD touch a circle at points P, Q, R and S respectively, then CR + PB =

1. BC

2. AB

3. CD

4. AD

Answer

Tangents drawn from the same external point to a circle are equal.

From B,

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

From C,

CR = CQ ..........(2)

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

∴ CR + PB = CQ + BQ = BC

Hence, option 1 is the correct option.

In the figure, sides MN, NL and LM of ΔLMN touch a circle at the points A, B and C respectively. If AN = 5 cm, CL = 4 cm and CM = 6 cm, then the perimeter of ΔLMN is :

1. 30 cm

2. 45 cm

3. 60 cm

4. 15 cm

Answer

Tangents drawn from the same external point to a circle are equal.

From M,

MA = CM = 6 cm

From N,

NA = NB = 5 cm

From L,

LB = LC = 4 cm

MN = MA + AN = 6 + 5 = 11

NL = NB + BL = 5 + 4 = 9

LM = LC + CM = 4 + 6 = 10

Perimeter of ΔLMN = MN + LN + LM

= 11 + 9 + 10

= 30 cm.

Hence, option 1 is the correct option.

In the figure, two circles touch each other at X. YZ and PX are common tangents to these circles. If YP = 3.8 cm, then YZ = ?

1. 1.9 cm

2. 11.4 cm

3. 7.6 cm

4. 7 cm

Answer

We know that if two tangents are drawn from an external point to a circle then, the lengths of the tangents are equal.

From figure,

PX and PY are the tangents to the first circle.

∴ PX = PY = 3.8 cm

PX and PZ are tangents to the second circle.

∴ PZ = PX = 3.8 cm

From figure,

YZ = PZ + PY = 3.8 + 3.8 = 7.6 cm.

Hence, option 3 is the correct option.

In the figure, AB is a chord of the circle such that ∠AXB = 50°. If AP is tangent to the circle at point A, then ∠BAP = ?

1. 65°

2. 50°

3. 40°

4. can’t be determined

Answer

From figure,

∠BAP = ∠AXB = 50° [Angle in alternate segment are equal]

Hence, option 2 is the correct option.

In the given figure, RT is a tangent touching the circle at S. If ∠PST = 30° and ∠SPQ = 60°, then ∠PSQ is :

1. 40°

2. 30°

3. 60°

4. 90°

Answer

We know that,

Angle between tangent and the chord at the point of contact is equal to angle of the alternate segment.

∴ ∠PQS = ∠PST = 30°

In △ PQS,

By angle sum property of triangle,

⇒ ∠PQS + ∠QPS + ∠PSQ = 180°

⇒ 30° + 60° + ∠PSQ = 180°

⇒ ∠PSQ + 90° = 180°

⇒ ∠PSQ = 180° - 90° = 90°.

Hence, option 4 is the correct option.

In the given figure, O is the centre of the circle and AB is a chord. If the tangent AM at A makes an angle of 50° with AB, then ∠AOB = ?

1. 100°

2. 75°

3. 80°

4. 150°

Answer

In a circle, radius through the point of contact is perpendicular to the tangent.

∠OAM = 90°

∠OAB + ∠BAM = 90°

∠OAB + 50° = 90°

∠OAB = 90° - 50°

∠OAB = 40°

OA = OB (Radii of same circle)

∠OBA = ∠OAB = 40° [Angles opposite to equal sides in a triangle are equal]

In △AOB,

By angle sum property of triangle,

⇒ ∠OBA + ∠OAB + ∠AOB = 180°

⇒ 40° + 40° + ∠AOB = 180°

⇒ ∠AOB + 80° = 180°

⇒ ∠AOB = 180° - 80° = 100°.

Hence, option 1 is the correct option.

In the figure, XY is a tangent at X to the circle with centre O. If ∠XYO = 25°, then x = ?

1. 25°

2. 115°

3. 65°

4. 60°

Answer

∠XYO = 25°

In a circle, radius through the point of contact is perpendicular to the tangent.

∠OXY = 90°

In △ XOY,

By angle sum property of triangle,

⇒ ∠XYO + ∠OXY + ∠XOY = 180°

⇒ 25° + 90° + ∠XOY = 180°

⇒ ∠XOY + 115° = 180°

⇒ ∠XOY = 180° - 115° = 65°.

From figure,

∠XOY + x = 180° [Linear pairs]

x = 180° - 65°

x = 115°.

Hence, option 2 is the correct option.

In the adjoining diagram, O is the center of the circle and PT is a tangent. The value of x is :

1. 20°

2. 40°

3. 55°

4. 70°

Answer

From figure,

⇒ ∠QOT + ∠TOP = 180° (Linear Pair)

⇒ 110° + ∠TOP = 180°

⇒ ∠TOP = 180° - 110° = 70°.

⇒ ∠OPT = 90° (Tangent is perpendicular to radius at point of intersection)

In △ TOP,

⇒ ∠OPT + ∠TOP + ∠PTO = 180°

⇒ 90° + 70° + x°= 180°

⇒ x° + 160° = 180°

⇒ x° = 180° - 160° = 20°.

Hence, option 1 is the correct option.

In the given figure, PT and QT are tangents to a circle such that ∠TPS = 45° and ∠TQS = 30°. Then, the value of x is :

1. 30°

2. 45°

3. 75°

4. 105°

Answer

We know that,

The angle between a tangent and a chord through point of contact is equal to an angle in the alternate segment.

∴ ∠SQP = ∠SPT = 45° and ∠SPQ = ∠SQT = 30°.

In △ SQP,

⇒ ∠SQP + ∠SPQ + ∠QSP = 180°

⇒ 45° + 30° + ∠QSP = 180°

⇒ 75° + ∠QSP = 180°

⇒ ∠QSP (x) = 180° - 75° = 105°.

Hence, option 4 is the correct option.

In the given figure, XQY is a tangent at Q to a circle. If PM is a chord parallel to XY and ∠MQY = 70°, then ∠PQM = ?

1. 20°

2. 35°

3. 40°

4. 70°

Answer

From figure,

MQ is chord and XQY is a tangent.

∠P = ∠MQY (∵ angles in alternate segment are equal.)

As PM || XQY

∠MQY = ∠M (∵ alternate angles are equal)

∴ ∠P = ∠M = 70°

We know that sum of angles in a triangle = 180°.

In △PQM,

⇒ ∠P + ∠M + ∠PQM = 180°

⇒ 70° + 70° + ∠PQM = 180°

⇒ 140° + ∠PQM = 180°

⇒ ∠PQM = 180° - 140°

⇒ ∠PQM = 40°.

Hence, option 3 is the correct option.

In the given diagram, O is the center of the circle, and PQ is a tangent at A. If ∠ABC = 50°, then values of x, y and z respectively are :

1. 50°, 100°, 40°

2. 50°, 50°, 65°

3. 40°, 80°, 50°

4. 50°, 25°, 78°

Answer

We know that,

The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

∴ ∠AOC (y) = 2∠ABC = 2 × 50° = 100°.

In △ AOC,

⇒ OA = OC (Radii of same circle)

⇒ ∠OAC = ∠OCA = z (Angle opposite to equal sides are equal)

By angle sum property of triangle,

⇒ ∠OAC + ∠OCA + ∠AOC = 180°

⇒ z + z + 100° = 180°

⇒ 2z = 180° - 100°

⇒ 2z = 80°

⇒ z = $\dfrac{80°}{2}$ = 40°.

We know that,

The tangent at any point of a circle is perpendicular to the radius through the point of contact.

⇒ ∠CAQ (x) = ∠OAQ - ∠OAC = 90° - 40° = 50°.

Hence, option 1 is the correct option.

In the given diagram, the radius of the circle with centre O is 3 cm. PA and PB are the tangents to the circle which are at right angle to each other. The length of OP is:

1. $\dfrac{3}{\sqrt{2}}$ cm

2. 3 cm

3. $3\sqrt{2}$ cm

4. $6\sqrt{2}$ cm

Answer

Join OA and AP.

Given,

Radius of circle (OB) = 3 cm

PB and AP are at right angles.

We know that,

Radius and tangent at point of contact are perpendicular to each other.

OA ⊥ AP

From figure,

⇒ OA = OB = 3 cm [Radius of circle]

⇒ OB is perpendicular to PB

⇒ ∠OBP = 90°

This shows APBO is square.

OA = PB = AP = OB = 3 cm.

In right angled triangle OBP,

⇒ OP² = OB² + PB²

⇒ OP² = 3² + 3²

⇒ OP² = 9 + 9

⇒ OP² = 18

⇒ OP = $\sqrt{18}$

⇒ OP = $3\sqrt{2}$ cm.

Hence, option 3 is the correct option.

In the adjoining diagram, PQ is a tangent at A to the circle with centre O. If ∠OAC = 25°, then ∠ABC is:

1. 20°

2. 65°

3. 70°

4. 130°

Answer

Given,

∠OAC = 25°

We know that,

Radius from the center and tangent are perpendicular to each other at point of contact.

⇒ ∠OAQ = ∠OAP = 90°

From figure,

⇒ ∠OAQ = ∠OAC + ∠CAQ

⇒ 90° = 25° + ∠CAQ

⇒ ∠CAQ = 90° - 25°

⇒ ∠CAQ = 65°.

We know that,

The angle between a tangent and a chord through the point of contact is equal to an angle in alternate segment. Hence,

⇒ ∠ABC = ∠CAQ

⇒ ∠ABC = 65°.

Hence, option 2 is the correct option.