The line 2x – y = 3 … (1)
The points (3, -2), (-2, 0)
The equation of the circle: x2 + y2 + 2ax + 2by + c = 0 ….. (2)
substituting the centre (-a, -b) in equation (1) we get,
\[\begin{array}{*{35}{l}}
2\left( -\text{ }a \right)\text{ }\text{ }-\left( -\text{ }b \right)\text{ }=\text{ }3 \\
-2a\text{ }+\text{ }b\text{ }=\text{ }3 \\
2a\text{ }-\text{ }b\text{ }+\text{ }3\text{ }=\text{ }0\ldots \ldots \text{ }\left( 3 \right) \\
\end{array}\]
Now Substitute the given points (3, -2) in equation (2), we get
\[\begin{array}{*{35}{l}}
{{3}^{2}}~+\text{ }{{\left( -\text{ }2 \right)}^{2}}~+\text{ }2a\left( 3 \right)\text{ }+\text{ }2b\left( -\text{ }2 \right)\text{ }+\text{ }c\text{ }=\text{ }0 \\
9\text{ }+\text{ }4\text{ }+\text{ }6a\text{ }-\text{ }4b\text{ }+\text{ }c\text{ }=\text{ }0 \\
6a\text{ }-\text{ }4b\text{ }+\text{ }c\text{ }+\text{ }13\text{ }=\text{ }0\ldots ..\text{ }\left( 4 \right) \\
\end{array}\]
Substitute the points (-2, 0) in equation (2), we get
\[\begin{array}{*{35}{l}}
{{\left( -\text{ }2 \right)}^{2}}~+\text{ }{{0}^{2}}~+\text{ }2a\left( -\text{ }2 \right)\text{ }+\text{ }2b\left( 0 \right)\text{ }+\text{ }c\text{ }=\text{ }0 \\
4\text{ }+\text{ }0\text{ }-\text{ }4a\text{ }+\text{ }c\text{ }=\text{ }0 \\
4a\text{ }-\text{ }c\text{ }-\text{ }4\text{ }=\text{ }0\ldots ..\text{ }\left( 5 \right) \\
\end{array}\]
By simplifying the equations (3), (4) and (5) we get,
a = 3/2, b = 6, c = 2
Again by substituting the values of a, b, c in (2), we get
\[{{x}^{2}}~+\text{ }{{y}^{2}}~+\text{ }2\text{ }\left( 3/2 \right)x\text{ }+\text{ }2\text{ }\left( 6 \right)y\text{ }+\text{ }2\text{ }=\text{ }0\]
\[{{x}^{2}}~+\text{ }{{y}^{2}}~+\text{ }3x\text{ }+\text{ }12y\text{ }+\text{ }2\text{ }=\text{ }0\]
∴ The equation of the circle is x2 + y2 + 3x + 12y + 2 = 0.