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