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

Let \((h, k) b\) the centre of the circle. It lies on the line \(x-4 y=1\)
\(
\begin{aligned}
& \Rightarrow \mathrm{h}=1+4 \mathrm{k} \\
& \therefore \text { centre } \equiv(4 \mathrm{k}+1, \mathrm{k})
\end{aligned}
\)
Circle passes through points \((3,7)\) and \((5,5)\)
\(\therefore(4 \mathrm{k}+1-5)^2+(\mathrm{k}-5)^2=(4 \mathrm{k}-2)^2+(\mathrm{k}-7)^2 \)
\( \therefore 16 \mathrm{k}^2+16-32 \mathrm{k}+\mathrm{k}^2+25-10 \mathrm{k}=16 \mathrm{k}^2+4-\)\(16 \mathrm{k}+\mathrm{k}^2+49 \)
\( -14 \mathrm{k}\)
\(\therefore-42 \mathrm{k}+41=-30 \mathrm{k}+53 \Rightarrow 12 \mathrm{k}=-12 \Rightarrow \mathrm{k}=\) \(-1 \)
\( \therefore \text { centre } \equiv(-3,-1) \)
\( \therefore \text { Radius }=\sqrt{(-3-5)^2+(-1-5)^2}=10\)
Hence equation of required circle is
\(
(x+3)^2+(y+1)^2=(10)^2 \text { i.e. } x^2+y^2+6 x+2 y-\)\(90=0
\)