PROBLEMS ON TANGENT DRAWN FROM AN EXTERNAL POINT TO A CIRCLE

Problem 1 :

If the angle between two tangents drawn from an external point 𝑃 to a circle of radius 𝑎 and center 𝑂, is 60°, then find the length of 𝑂𝑃.

Solution :

In triangle OAP, using trigonometric ratio

sin θ = Opposite side/Hypotenuse

sin 30° = OA / OP

1/2 = a/OP

OP = 2a

So, length of OP is 2a.

Problem 2 :

𝑃𝑄 is a tangent drawn from an external point 𝑃 to a circle with center 𝑂, 𝑄𝑂𝑅 is the diameter of the circle. If ∠ 𝑃𝑂𝑅 = 120°, what is the measure of ∠𝑂𝑃𝑄?

Solution :

In triangle PQO,

∠OQP = 90

∠POQ + ∠POR = 180

∠POQ + 120 = 180

∠POQ = 180 - 120

∠POQ = 60

∠OPQ = 180 - (90 + 60)

∠OPQ = 180 - 150

∠OPQ = 30

Problem 3 :

In the given figure 𝑃𝐴 and 𝑃𝐵 are tangents to a circle with center 𝑂. If ∠𝐴𝑃𝐵 = (2𝑥 + 3)° and ∠𝐴𝑂𝐵 = (3𝑥 + 7)°, then find the value of 𝑥.

Solution :

∠APB + ∠BOA = 180

2x + 3 + 3x + 7 = 180

5x + 10 = 180

5x = 180 -10

5x = 170

x = 170/5

x = 34

Problem 4 :

In figure, 𝑃𝑄 is a tangent at a point 𝐶 to a circle with center 𝑂. If 𝐴𝐵 is a diameter and ∠𝐶𝐴𝐵 = 30°, find ∠𝑃𝐶𝐴.

Solution :

In triangle ACB,

∠𝐶𝐴𝐵 = 30, ∠ACB = 90, ∠ABC = ?

∠𝐶𝐴𝐵 + ∠ACB + ∠ABC = 180

30 + 90 + ∠ABC = 180

120 + ∠ABC = 180

∠ABC = 180 - 120

∠ABC = 60

∠PCA = 60 (using alternate segment theorem)

Problem 5 :

In figure, 𝐴𝑂𝐵 is a diameter of a circle with centre 𝑂 and 𝐴𝐶 is a tangent to the circle at 𝐴. If ∠𝐵𝑂𝐶 = 130°, then find ∠𝐴𝐶𝑂.

Solution :

∠BOC + ∠COA = 180

130 + ∠COA = 180

∠COA = 180 - 130

∠COA = 50

In triangle AOC,

∠OAC + ∠ACO + ∠COA = 180.

90 + ∠ACO + 50 = 180

∠ACO + 140 = 180

∠ACO = 180 - 140

∠ACO = 40

Problem 6 :

In figure, 𝑃𝐴 and 𝑃𝐵 are tangents to the circle with center 𝑂 such that ∠𝐴𝑃𝐵 = 50°. Write the measure of ∠𝑂𝐴B

Solution :

∠PAB = ∠PBA

Let ∠PBA = x

50 + x + x = 180

2x = 180 - 50

2x = 130

x = 65

∠PAO = 90

∠PAB + ∠BAO = 90

65 + ∠BAO = 90

∠BAO = 90 - 65

∠BAO = 25

Problem 7 :

In figure, 𝑃𝐴 and 𝑃𝐵 are two tangents drawn from an external point 𝑃 to a circle with center 𝐶 and radius 4 𝑐𝑚. If 𝑃𝐴 ⊥ 𝑃𝐵, then the length of each tangent is:

Solution :

Triangle ACP is 45-45-90 special right triangle.

∠ACP = ∠APC

AC = AP = 4 cm

Problem 8 :

In figure, the sides 𝐴𝐵,𝐵𝐶 and 𝐶𝐴 of a triangle 𝐴𝐵𝐶, touch a circle at 𝑃, 𝑄 and 𝑅 respectively. If 𝑃𝐴 = 4 𝑐𝑚, 𝐵𝑃 = 3 𝑐𝑚 and 𝐴𝐶 = 11 𝑐𝑚, then the length of 𝐵𝐶 (in 𝑐𝑚) is:

Solution :

 𝑃𝐴 = 4 𝑐𝑚, 𝐵𝑃 = 3 𝑐𝑚 and 𝐴𝐶 = 11 𝑐𝑚

AP = AR = 4 cm

BP = BQ = 3 cm

AB = AP + PB

AB = 4 + 3

AB = 7 cm

AC = 11 cm

AR + RC = 11

4 + RC = 11

RC = 7 cm

CQ = 7 cm

Then,

BC = BQ + QC

BC = 3 + 7

BC = 10 cm

Problem 9 :

In figure, a circle touches the side 𝐷𝐹 of Δ𝐸𝐷𝐹 at 𝐻 and touches 𝐸𝐷 and 𝐸𝐹 produced at 𝐾 and 𝑀 respectively. If 𝐸𝐾 = 9 𝑐𝑚, then the perimeter of Δ𝐸𝐷𝐹 (in 𝑐𝑚) is:

Solution :

EK = 9 cm

DH = DK and FH = FM

In triangle EDF,

Perimeter of triangle EDF :

= ED + DF + EF

= (ED + DH) + (HF + EF)

= 9 + 9

= 18 cm

Problem 10 :

In figure, 𝑃𝑄 and 𝑃𝑅 are tangents to a circle with center 𝐴. If ∠𝑄𝑃𝐴 = 27°, then ∠𝑄𝐴𝑅 equals.

Solution :

In triangle QAP,

∠𝑄𝐴P = 180 - 90 - 27

∠𝑄𝐴P = 63

∠𝑄𝐴P = ∠PAR

∠𝑄𝐴R = 2(63)

∠𝑄𝐴R = 126

Problem 11 :

The length of a tangent PQ from external point P is 24 cm. If the distance of the point P from the center is 25 cm, then the diameter of the circle is 

a) 15 cm       b) 14 cm       c) 7 cm     d)  12 cm

Solution :

A line which is drawn from the center of the circle to the point of tangency will be perpendicular.

The distance between the center and the external point is the hypotenuse.

Let x be the radius.

25² = x² + 24²

625 = x² + 576

x² = 625 - 576

x² = 49

x = 7

Diameter = 2(7)

= 14 cm

So, option b is correct.