The points (3, 7), (5, 5)
The line x – 4y = 1…. (1)
the equation of the circle: x2 + y2 + 2ax + 2by + c = 0 ….. (2)
substituting the centre (-a, -b) in equation (1) we get,
\[\begin{array}{*{35}{l}}
\left( -\text{ }a \right)\text{ }-\text{ }4\left( -\text{ }b \right)\text{ }=\text{ }1 \\
-a\text{ }+\text{ }4b\text{ }=\text{ }1 \\
a\text{ }-\text{ }4b\text{ }+\text{ }1\text{ }=\text{ }0\ldots \ldots \text{ }\left( 3 \right) \\
\end{array}\]
Substituting the points (3, 7) in equation (2), we get
\[\begin{array}{*{35}{l}}
{{3}^{2}}~+\text{ }{{7}^{2}}~+\text{ }2a\left( 3 \right)\text{ }+\text{ }2b\left( 7 \right)\text{ }+\text{ }c\text{ }=\text{ }0 \\
9\text{ }+\text{ }49\text{ }+\text{ }6a\text{ }+\text{ }14b\text{ }+\text{ }c\text{ }=\text{ }0 \\
6a\text{ }+\text{ }14b\text{ }+\text{ }c\text{ }+\text{ }58\text{ }=\text{ }0\ldots ..\text{ }\left( 4 \right) \\
\end{array}\]
Substituting the points (5, 5) in equation (2), we get
\[\begin{array}{*{35}{l}}
{{5}^{2}}~+\text{ }{{5}^{2}}~+\text{ }2a\left( 5 \right)\text{ }+\text{ }2b\left( 5 \right)\text{ }+\text{ }c\text{ }=\text{ }0 \\
25\text{ }+\text{ }25\text{ }+\text{ }10a\text{ }+\text{ }10b\text{ }+\text{ }c\text{ }=\text{ }0 \\
10a\text{ }+\text{ }10b\text{ }+\text{ }c\text{ }+\text{ }50\text{ }=\text{ }0\ldots ..\text{ }\left( 5 \right) \\
\end{array}\]
By simplifying equations (3), (4) and (5) we get,
a = 3, b = 1, c = – 90
Now, by substituting the values of a, b, c in equation (2), we get
\[\begin{array}{*{35}{l}}
{{x}^{2}}~+\text{ }{{y}^{2}}~+\text{ }2\left( 3 \right)x\text{ }+\text{ }2\left( 1 \right)y\text{ }\text{ }90\text{ }=\text{ }0 \\
{{x}^{2}}~+\text{ }{{y}^{2}}~+\text{ }6x\text{ }+\text{ }2y\text{ }-\text{ }90\text{ }=\text{ }0 \\
\end{array}\]
∴ The equation of the circle is x2 + y2 + 6x + 2y – 90 = 0.