$\left\{ \begin{array}{l}
x^2 +y^2 -4x -22y +75 = 0 & \ldots \unicode{x2460} \\
4x=3y & \ldots \unicode{x2461}
\end{array}\right.$
From $\unicode{x2461}$, we have
$\begin{array}{rcl}
4x & = & 3y \\
x & = & \dfrac{3y}{4} \ldots \unicode{x2462}
\end{array}$
Sub. $\unicode{x2462}$ into $\unicode{x2460}$, we have
$\begin{array}{rcl}
\left(\dfrac{3y}{4}\right)^2 +y^2 -4\left(\dfrac{3y}{4}\right) -22y +75 & = & 0 \\
\dfrac{9y^2}{16} +y^2 -3y -22y +75 & = & 0 \\
9y^2 +16y^2 -48y -352y +1200 & = & 0 \\
25y^2 -400y +1200 & = & 0 \\
y^2 -16y +48 & = & 0 \\
(y-4)(y-12) & = & 0 \\
\end{array}$
$\therefore y=4$ or $y=12$.
Sub. $y=4$ into $\unicode{x2462}$, we have
$\begin{array}{rcl}
x & = & \dfrac{3 \times 4}{4} \\
x & = & 3
\end{array}$
Sub. $y=12$ into $\unicode{x2462}$, we have
$\begin{array}{rcl}
x & = & \dfrac{3 \times 12}{4} \\
x & = & 9
\end{array}$
$\therefore$ the coordinates of the intersection points are $(3,4)$ and $(9,12)$.
The centre of the required circle
$\begin{array}{cl}
= & \left( \dfrac{3+9}{2}, \dfrac{4+12}{2} \right) \\
= & (6, 8)
\end{array}$
The radius of the required circle
$\begin{array}{cl}
= & \dfrac{1}{2} \sqrt{(9-3)^2+(12-4)^2} \\
= & 5
\end{array}$
Hence, the equation of the required circle is $(x-6)^2+(y-8)^2 =25$.
