JEE Main 2025 Jan 28 Shift 1, Mathematics Q8: Circle: Chord, Tangent And Normal

JEE Main2025Jan 28, Shift 1Mathematics

Let the equation of the circle, which touches x-axis at the point $(a, 0)$ , $a > 0$ and cuts off an intercept of length $b$ on 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 :

A
$(α, β^{2} + 4 γ)$
B
$(γ, β^{2} - 4 α)$
C
$(γ, β^{2} + 4 α)$
D
$(α, β^{2} - 4 γ)$

Solution

The circle touches the x-axis at $(a, 0)$ from below, so its centre is directly below the point of tangency at $(a, - r)$ , where $r$ is the radius.

Standard form: $(x - a)^{2} + (y + r)^{2} = r^{2}$ , i.e. $x^{2} + y^{2} - 2 a x + 2 r y + a^{2} = 0$ .

Comparing with $x^{2} + y^{2} - α x + β y + γ = 0$ :

$α = 2 a, β = 2 r, γ = a^{2}$ . So $2 a = α$ .

The chord cut on the y-axis has length $b$ . Putting $x = 0$ in the circle equation gives $y^{2} + β y + γ = 0$ . The length of the intercept equals $β^{2} - 4 γ$ , so $b^{2} = β^{2} - 4 γ$ .

$∴ (2 a, b^{2}) = (α, β^{2} - 4 γ)$ .

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JEE Main 2025 Jan 28 Shift 1, Mathematics Q8: Circle: Chord, Tangent And Normal

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