Two tangents are drawn from a point $P$ to the circle $x^{2}+y^{2}-2 x-4 y+4=0$, such that the angle between these tangents is $\tan ^{-1}\left(\frac{12}{5}\right)$, where $\tan ^{-1}\left(\frac{12}{5}\right) \in(0, \pi)$. If the centre of the circle is denoted by $C$ and these tangents touch the circle at points $A$ and $B$, then the ratio of the areas of $\triangle P A B$ and $\triangle C A B$ is:

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Circle

Two tangents are drawn from a point

P

to the circle

x^{2} + y^{2} - 2 x - 4 y + 4 = 0

, such that the angle between these tangents is

tan^{- 1} (\frac{12}{5})

, where

tan^{- 1} (\frac{12}{5}) \in (0, π)

. If the centre of the circle is denoted by

C

and these tangents touch the circle at points

A

and

B

, then the ratio of the areas of

△ P A B

and

△ C A B

is:

Answer: (A)

The circle is

(x - 1)^{2} + (y - 2)^{2} = 1

, with centre

C

and radius

1

. Let

∠ A P B = 2 θ

, so

tan 2 θ = \frac{12}{5}

and

\frac{2 tan θ}{1 - tan ^{2} θ} = \frac{12}{5} ⟹ 6 tan^{2} θ + 5 tan θ - 6 = 0

Since

tan 2 θ > 0

the angle

2 θ

is acute, so

θ

is acute and

tan θ = \frac{2}{3}

, giving

sin θ = \frac{2}{13}

and

cos θ = \frac{3}{13}

.
In the right triangle

P A C

sin θ = \frac{C A}{P C} = \frac{1}{P C}

, so

P C = \frac{13}{2}

and

P A = P C cos θ = \frac{3}{2}

.
The chord of contact

A B

is perpendicular to

P C

; let

M

be their intersection. Then

P M = P A cos θ = \frac{9}{2 13}, C M = C A sin θ = \frac{2}{13}

since the angle of triangle

P A C

C

\frac{π}{2} - θ

. Both triangles stand on the same base

A B

with heights

P M

and

C M

, so

\frac{area of △ P A B}{area of △ C A B} = \frac{P M}{C M} = \frac{9}{4}

and the ratio is

9 : 4

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Single correct option