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 (2a, b²) is equal to :

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$(α, β^{2} - 4 γ)$

$(α, β^{2} + 4 γ)$

$(γ, β^{2} + 4 α)$

$(γ, β^{2} - 4 α)$

answer is D.

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Detailed Solution

By pytogorus $r^{2} = a^{2} + \frac{b^{2}}{4} = P^{2}$

$r = \sqrt{\frac{4 a^{2} + b^{2}}{4}}$

Equation of circle is $(x - α)^{2} + (y - β)^{2} = r^{2}$

$x^{2} + y^{2} - 2 a x - 2 p y + α^{2} + p^{2} - r^{2} = 0$

comparision $x^{2} + y^{2} - α x + β y + r = 0$

$\begin{array}{l} - α = - 2 a, β = - 2 p, r = a^{2} \\ \Rightarrow 2 a = α, 4 a^{2} + b^{2} = 4 p^{2} \\ α^{2} + b^{2} = 4 p^{2} \\ α^{2} + b^{2} = β^{2} \end{array}$

So, $(2 a, b^{2}) = (α, β^{2} - 4 r)$

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