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

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