Let the shortest distance from $(a, 0), a > 0,$ to the parabola $y^{2} = 4 x$ be 4. Then the equation of the circle passing through the point $(a, 0)$ and the focus of the parabola, and having its centre on the axis of the parabola is

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$x^{2} + y^{2} - 6 x + 5 = 0$

$x^{2} + y^{2} - 4 x + 3 = 0$

$x^{2} + y^{2} - 10 x + 9 = 0$

$x^{2} + y^{2} - 8 x + 7 = 0$

answer is A.

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Detailed Solution

Normal at P
$\begin{matrix} y + tx = 2 t + t^{3} \\ ↑ \\ (a, 0) \end{matrix}$
Question Image
$\begin{array}{l} at = 2 t + t^{3} \\ a = 2 + t^{2} \\ ℝ (2 + t^{2}, 0) \\ P ℝ = 4 \Rightarrow 4 + 4 t^{2} = 16 \\ 4 t^{2} = 12 \Rightarrow t^{2} = 3 \\ a = 5 ℝ (5, 0) \end{array}$
Focus (1, 0) (1, 0) & (5, 0) will be the end pts. of diameter
$\begin{array}{l} \Rightarrow {Eg}^{n} of circle is \\ \begin{aligned} (x - 1) (x - 5) + y^{2} = 0 \\ x^{2} + y^{2} - 6 x + 5 = 0 \end{aligned} \end{array}$