A circle $\mathrm{C}_{1}$ passes through the origin O and has diameter 4 on the positive x -axis. The line $\mathrm{y}=2 \mathrm{x}$ gives a chord OA of a circle $\mathrm{C}_{1}$. Let $\mathrm{C}_{2}$ be the circle with OA as a diameter. If the tangent to $\mathrm{C}_{2}$ at the point A meets the x -axis at P and y -axis at Q , then $\mathrm{QA}: \mathrm{AP}$ is equal to :

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Circle

A circle

C_{1}

passes through the origin O and has diameter 4 on the positive x -axis. The line

y = 2 x

gives a chord OA of a circle

C_{1}

. Let

C_{2}

be the circle with OA as a diameter. If the tangent to

C_{2}

at the point A meets the x -axis at P and y -axis at Q , then

QA : AP

is equal to :

Answer: (A)

Circle

C_{1}

has equation

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

(center at

(2, 0)

, radius 2).

The line

y = 2 x

intersects

C_{1}

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

, giving

5 x^{2} - 4 x = 0

, so

x = 0

x = \frac{4}{5}

. Thus

A = (\frac{4}{5}, \frac{8}{5})

.

Circle

C_{2}

has OA as diameter. The tangent at A to

C_{2}

is perpendicular to OA.

Slope of OA is

tan θ = 2

. The tangent at A has slope

- \frac{1}{2}

(perpendicular).

Let

O A = ℓ

. Then:

\frac{Q A}{A P} = \frac{ℓ cot θ}{ℓ tan θ} = \frac{1}{tan ^{2} θ} = \frac{1}{4}

Therefore,

Q A : A P = 1 : 4

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Single correct option