The equation of the circle whose centre lies on the line \(x-4 y=1\) and which passes through the points \((3,7)\) and \((5,5)\) is

Let the equation of the circle be \(x^2+y^2+2 g x+2 f y+c=0\)
\(\because\) its centre lies on \(\mathrm{x}-4 \mathrm{y}=1\)
\(\Rightarrow-\mathrm{g}-4(-\mathrm{f})=1\)
\(\Rightarrow-g+4 f=1\quad\ldots\text{(i)}\)
\(\because\) it passes through \((3,7)\)
\(\Rightarrow 3^2+7^2+2 \mathrm{~g} \times 3+2 \mathrm{f} \times 7+\mathrm{c}=0\)
\(\Rightarrow 6 g+14 f+c=-58\quad\ldots\text{(ii)}\)
Also it passes through \((5,5)\)
\(\Rightarrow 5^2+5^2+2 g \times 5+2 f \times 5+c=0\)
\(\Rightarrow 10 g+10 f+c=-50\quad\ldots\text{(iii)}\)
from (iii) - (ii)
\(4 g-4 f=8 \Rightarrow g-f=2\)
\(
\begin{aligned}
& \Rightarrow \mathrm{f}=1 \\
& \Rightarrow \mathrm{g}=3 \\
& \Rightarrow \mathrm{c}=-90
\end{aligned}
\)
So, the equation of required circle is \(x^2+y^2+6 x+2 y-90=0\)