If the tangents drawn at the point $\mathrm{O}(0,0)$ and $P(1+\sqrt{5}, 2)$ on the circle $x^{2}+y^{2}-2 x-4 y=0$ intersect at the point Q , then the area of the triangle OPQ is equal to

JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Circle

If the tangents drawn at the point

O (0, 0)

and

P (1 + 5, 2)

on the circle

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

intersect at the point Q , then the area of the triangle OPQ is equal to

Answer: (C)

The tangent at

O (0, 0)

to the circle

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

- (x + 0) - 2 (y + 0) = 0

, i.e.,

x + 2 y = 0

.

The tangent at

P (1 + 5, 2)

x (1 + 5) + 2 y - (x + 1 + 5) - 2 (y + 2) = 0

. Substituting

x = - 2 y

from the first tangent and solving gives

y = - \frac{5 + 1}{2}

, so

Q (5 + 1, - \frac{5 + 1}{2})

.

The length of tangent from

O

O Q = \frac{5 + 5}{2}

. The radius of the circle is

R = 5

. Using the formula for the area of a triangle formed by two tangents from an external point: Area

= \frac{R L ^{3}}{R ^{2} + L ^{2}}

, where

L = O Q

, we compute the area as

\frac{5 + 3 5}{2}

JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Single correct option