A Circle touching the x-axis at ( 3, 0 ) and making an intercept of length 8 on the y-axis passes through the point

$\begin{array}{l}{x}^2+y^2+2gx+2fy+c=0\\ \text{passing through\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}(3,0)\\ 9+0+6g+c=0\\ 6g+c+9=0\\ g^2=c\end{array}$

$\begin{array}{l}2\sqrt{f^2-c}=8\\ \sqrt{f^2-c}=4\\ f^2-c=16\end{array}$

$\begin{array}{l}6g+g^2+9=0 9=f^2+16\\ {(g+3)}^2=0 f^2=9+16\\ g=-3 f=\pm 5\\ c={(-3)}^2 c=(-g,-f)\\ c=9 c=(3,-(\pm 5))\end{array}$

C=(3, -5) or c= (3, 5)

The eq.of circle
$\begin{array}{l}{(x-3)}^2+{(y-5)}^2=5^2\\ {(x-3)}^2+{(y+5)}^2=5^2\end{array}$
Hence (3, 10) will satisfy equation (1)