How to find the equation of circle whose diameter is the latus rectum of the parabola.

Question

The only hint given in this question is $x^2 = -36 y$

I am having problems starting the question I am clueless how to solve it.

Answer 1

The coordinate of the focus of $x^2=-4ay$ is $(0,-a)$

So, the equation of the latus rectum will be $y=-a$ as it is perpendicular to the axis of the Parabola $x=0$

So, putting $y=-a$ in the equation of the Parabola $x^2=-4a(-a)\implies x=\pm2a$

So, the coordinates of the two end points of the latus rectum are $(\pm 2a,-a)$

Then the centre of the circle will be $(\frac{2a-2a}2,\frac{-a-a}2)$ or $(0,-a)$ which happens to be the focus of the Parabola.

Can you take it from here?