Normal at $P (2, 4)$ to $y^{2} = 8 x$ meets the parabola at $Q$ . Then the equation of the circle on normal chord $P Q$ as diameter is

x^{2} + y^{2} - 20 x + 8 y - 12 = 0

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x^{2} + y^{2} - 10 x + 4 y - 8 = 0

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x^{2} + y^{2} - 12 x + 6 y - 15 = 0

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x^{2} + y^{2} - 10 x + 8 y - 12 = 0

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Solution

The correct option is B $x^{2} + y^{2} - 20 x + 8 y - 12 = 0$
Normal at $(a t_{1}^{2}, 2 a t_{1})$ meets the parabola again at $(a t_{2}^{2}, 2 a t_{2})$
Then,
$t_{2} = - t_{1} - \frac{2}{t_{1}}$
Given parabola is $y^{2} = 8 x$
$\Rightarrow a = 2$
$P = (2, 4)$
$\Rightarrow 2 a t_{1} = 4$
$∴ t_{1} = 1$
$\Rightarrow t_{2} = - t_{1} - \frac{2}{t_{1}}$
$\Rightarrow t_{2} = - 3$
Then, $P = (2, 4)$ and $Q = (18, - 12)$
Then, equation of circle is given by
$(x - 2) (x - 18) + (y - 4) (y + 12) = 0$
$∴ x^{2} + y^{2} - 20 x + 8 y - 12 = 0$

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