Q. Let $P$ be the point on the parabola, $y^2=8x$ which is at a minimum distance from the centre $C$ of the circle, $x^2 + (y+6)^2=1$. Then the equation of the circle, passing through $C$ and having its centre at $P$ is :
Solution:
Normal at $P (at^2 , 2at) $ is $y + tx = 2at + at^3$
Given it passes $(0, -6)$
$\Rightarrow \, - 6 = 2at + at^3 (a = 2)$
$ - 6 = 4t + 2t^3$
$t^3 + 2t + 3 = 0 $
$t = - 1 $
so, $P (a, -2a) = (2, -4) . [a = 1)$
radius of circle $= CP = \sqrt{2^2 + (-4 + 6)^2} = 2 \sqrt{2}$
Circle is $(x - 2)^2 + (y + 4)^2 = (2 \sqrt{2})^2$
$x^2 + y^2 - 4x + 8y + 12 = 0 $
