Answer: \((x-4)^2+(y-2)^2=40\), and \((0,\,2\pm2\sqrt6)\).

(a) The radius is the distance from the centre \((4,2)\) to the point \((-2,0)\) on the circle.

So

\(r^2=(-2-4)^2+(0-2)^2.\)

Thus

\(r^2=36+4=40.\)

The equation of a circle with centre \((4,2)\) and radius squared \(40\) is

\((x-4)^2+(y-2)^2=40.\)

(b) Points on the \(y\)-axis have \(x=0\). Substitute \(x=0\) into the circle equation:

\((0-4)^2+(y-2)^2=40.\)

So

\(16+(y-2)^2=40.\)

Hence

\((y-2)^2=24.\)

Taking square roots,

\(y-2=\pm\sqrt{24}=\pm2\sqrt6.\)

Therefore

\(y=2\pm2\sqrt6.\)

The points where the circle meets the \(y\)-axis are

\((0,2+2\sqrt6)\quad\text{and}\quad(0,2-2\sqrt6).\)