Let the equation of the circle, which touches x -axis at the point ( $\mathrm{a}, 0$ ), $\mathrm{a}>0$ and cuts off an intercept of length $b$ on $y$-axis be $x^{2}+y^{2}-\alpha x+\beta y+\gamma=0$. If the circle lies below x -axis, then the ordered pair $\left(2 \mathrm{a}, \mathrm{b}^{2}\right)$ is equal to :

(D) $\left(\alpha, \beta^{2}-4 \gamma\right)$

(A) $\left(\alpha, \beta^{2}+4 \gamma\right)$

(B) $\left(\gamma, \beta^{2}-4 \alpha\right)$

(C) $\left(\gamma, \beta^{2}+4 \alpha\right)$

JEE Main · 2025 · 28 Jan 2025 · Shift 1 · Circle

Let the equation of the circle, which touches x -axis at the point (

a, 0

a > 0

and cuts off an intercept of length

b

y

-axis be

x^{2} + y^{2} - α x + β y + γ = 0

. If the circle lies below x -axis, then the ordered pair

(2 a, b^{2})

is equal to :

Answer: (D)

The circle touches the x-axis at

(a, 0)

and lies below the x-axis, so its centre is

(a, - r)

, where

r

is the radius.

Equation:

(x - a)^{2} + (y + r)^{2} = r^{2} \Rightarrow x^{2} + y^{2} - 2 a x + 2 r y + a^{2} = 0

Comparing with

x^{2} + y^{2} - α x + β y + γ = 0

α = 2 a, β = 2 r, γ = a^{2}

.

Setting

x = 0

gives the y-intercepts:

y^{2} + β y + γ = 0

, whose roots differ by

b = β^{2} - 4 γ

, so

b^{2} = β^{2} - 4 γ

Therefore

(2 a, b^{2}) = (α, β^{2} - 4 γ)

JEE Main · 2025 · 28 Jan 2025 · Shift 1 · Single correct option