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

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

Answer: gradients \(AB=\frac34\), \(AC=-\frac17\), \(BC=-\frac43\); circle \((x-5)^2+(y-3)^2=50\).

(a) Use \(m=\dfrac{y_2-y_1}{x_2-x_1}\).

For \(AB\),

\(m_{AB}=\dfrac{10-4}{6-(-2)}=\dfrac68=\dfrac34.\)

For \(AC\),

\(m_{AC}=\dfrac{2-4}{12-(-2)}=-\dfrac{2}{14}=-\dfrac17.\)

For \(BC\),

\(m_{BC}=\dfrac{2-10}{12-6}=-\dfrac86=-\dfrac43.\)

(b) Since

\(m_{AB}\,m_{BC}=\dfrac34\left(-\dfrac43\right)=-1,\)

the lines \(AB\) and \(BC\) are perpendicular. Therefore \(\angle ABC=90^\circ\).

By the angle in a semicircle result, if \(ABC\) is a right angle and the circle passes through \(A\), \(B\) and \(C\), then \(AC\) is a diameter of the circle.

The centre is the midpoint of \(A(-2,4)\) and \(C(12,2)\):

\(\left(\dfrac{-2+12}{2},\dfrac{4+2}{2}\right)=(5,3).\)

The radius squared can be found using \(A\):

\(r^2=(-2-5)^2+(4-3)^2=49+1=50.\)

Therefore the equation of the circle is

\((x-5)^2+(y-3)^2=50.\)