Equation of a Circle and Geometry Question

A circle has equation $x^2+y^2+4x-6y=36$
(a) Find the centre and radius of the circle.
Points $S$ and $T$ lie on the circle such that the origin is the midpoint of $ST$ .
(b) Show that $ST$ has a length of 12.

(a)We need to put the circle equation into completed square form

$\begin{equation*}(x+2)^2-4+(y-3)^2-9=36\end{equation}$

$\begin{equation*}(x+2)^2+(y-3)^2=49\end{equation}$

The centre is $(-2, 3)$ and the radius is $7$ .

(b)Draw a diagram

We know $SO$ and $TO$ are radii of the circle. Hence $\Delta{SOT}$ is isosceles and the line segment from $O$ to the origin is perpendicular to $ST$ .