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$The tangent and normal at the point A (4, 4) to the parabola y^{2} = 4 x intersect x -axis at the$
$points B and C respectively, then the equation of circumcircle of △ ABC is$

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$x^{2} + y^{2} - 4 x - 6 y = 0$

$x^{2} + y^{2} - 4 x + 6 y = 0$

$x^{2} + y^{2} + 4 x - 6 y = 0$

$x^{2} + y^{2} - 2 x - 24 = 0$

answer is D.

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Detailed Solution

$B = (- 4, 0), C = (6, 0) are ends of diameter of circle.$

$eq. of circle is (x + 4) (x - 6) + y^{2} = 0$

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