(i) The lines \[\begin{array}{*{35}{l}}
x\text{ }+\text{ }y\text{ }+\text{ }3\text{ }=\text{ }0 \\
x\text{ }-\text{ }y\text{ }+\text{ }1\text{ }=\text{ }0 \\
x\text{ }=\text{ }3 \\
\end{array}\]
solving these lines we get the intersection points A (-2, -1), B (3, 4), C (3, -6)
the equation of the circle: x2 + y2 + 2ax + 2by + c = 0….. (1)
Substitute the points (-2, -1) in equation (1), we get
\[\begin{array}{*{35}{l}}
{{\left( -\text{ }2 \right)}^{2}}~+\text{ }{{\left( -\text{ }1 \right)}^{2}}~+\text{ }2a\left( -2 \right)\text{ }+\text{ }2b\left( -1 \right)\text{ }+\text{ }c\text{ }=\text{ }0 \\
4\text{ }+\text{ }1\text{ }-\text{ }4a\text{ }-\text{ }2b\text{ }+\text{ }c\text{ }=\text{ }0 \\
5\text{ }-\text{ }4a\text{ }-\text{ }2b\text{ }+\text{ }c\text{ }=\text{ }0 \\
4a\text{ }+\text{ }2b\text{ }-\text{ }c\text{ }-\text{ }5\text{ }=\text{ }0\ldots ..\text{ }\left( 2 \right) \\
\end{array}\]
Substitute the points (3, 4) in equation (1), we get
\[\begin{array}{*{35}{l}}
{{3}^{2}}~+\text{ }{{4}^{2}}~+\text{ }2a\left( 3 \right)\text{ }+\text{ }2b\left( 4 \right)\text{ }+\text{ }c\text{ }=\text{ }0 \\
9\text{ }+\text{ }16\text{ }+\text{ }6a\text{ }+\text{ }8b\text{ }+\text{ }c\text{ }=\text{ }0 \\
6a\text{ }+\text{ }8b\text{ }+\text{ }c\text{ }+\text{ }25\text{ }=\text{ }0\ldots ..\text{ }\left( 3 \right) \\
\end{array}\]
Substitute the points (3, -6) in equation (1), we get
\[\begin{array}{*{35}{l}}
{{3}^{2}}~+\text{ }{{\left( -\text{ }6 \right)}^{2}}~+\text{ }2a\left( 3 \right)\text{ }+\text{ }2b\left( -\text{ }6 \right)\text{ }+\text{ }c\text{ }=\text{ }0 \\
9\text{ }+\text{ }36\text{ }+\text{ }6a\text{ }-\text{ }12b\text{ }+\text{ }c\text{ }=\text{ }0 \\
6a\text{ }-\text{ }12b\text{ }+\text{ }c\text{ }+\text{ }45\text{ }=\text{ }0\ldots ..\text{ }\left( 4 \right) \\
\end{array}\]
Upon simplifying equations (2), (3), (4) we get
a = – 3, b = 1, c = -15.
Now by substituting the values of a, b, c in equation (1), we get
\[\begin{array}{*{35}{l}}
{{x}^{2}}~+\text{ }{{y}^{2}}~+\text{ }2\left( -\text{ }3 \right)x\text{ }+\text{ }2\left( 1 \right)y\text{ }-\text{ }15\text{ }=\text{ }0 \\
{{x}^{2}}~+\text{ }{{y}^{2}}~-\text{ }6x\text{ }+\text{ }2y\text{ }\text{ }15\text{ }=\text{ }0 \\
\end{array}\]
∴ The equation of the circle is x2 + y2 – 6x + 2y – 15 = 0.
(ii) The lines \[\begin{array}{*{35}{l}}
2x\text{ }+\text{ }y\text{ }-\text{ }3\text{ }=\text{ }0 \\
x\text{ }+\text{ }y\text{ }-\text{ }1\text{ }=\text{ }0 \\
3x\text{ }+\text{ }2y\text{ }-\text{ }5\text{ }=\text{ }0 \\
\end{array}\]
On solving, the intersection points A(2, – 1), B(3, – 2), C(1,1)
the equation of the circle: x2 + y2 + 2ax + 2by + c = 0….. (1)
Substitute the points (2, -1) in equation (1), we get
\[\begin{array}{*{35}{l}}
{{2}^{2}}~+\text{ }{{\left( -\text{ }1 \right)}^{2}}~+\text{ }2a\left( 2 \right)\text{ }+\text{ }2b\left( -\text{ }1 \right)\text{ }+\text{ }c\text{ }=\text{ }0 \\
4\text{ }+\text{ }1\text{ }+\text{ }4a\text{ }-\text{ }2b\text{ }+\text{ }c\text{ }=\text{ }0 \\
4a\text{ }-\text{ }2b\text{ }+\text{ }c\text{ }+\text{ }5\text{ }=\text{ }0\ldots ..\text{ }\left( 2 \right) \\
\end{array}\]
Substitute the points (3, -2) in equation (1), we get
\[\begin{array}{*{35}{l}}
{{3}^{2}}~+\text{ }{{\left( -\text{ }2 \right)}^{2}}~+\text{ }2a\left( 3 \right)\text{ }+\text{ }2b\left( -\text{ }2 \right)\text{ }+\text{ }c\text{ }=\text{ }0 \\
9\text{ }+\text{ }4\text{ }+\text{ }6a\text{ }-\text{ }4b\text{ }+\text{ }c\text{ }=\text{ }0 \\
6a\text{ }-\text{ }4b\text{ }+\text{ }c\text{ }+\text{ }13\text{ }=\text{ }0\ldots ..\text{ }\left( 3 \right) \\
\end{array}\]
Substitute the points (1, 1) in equation (1), we get
\[\begin{array}{*{35}{l}}
{{1}^{2}}~+\text{ }{{1}^{2}}~+\text{ }2a\left( 1 \right)\text{ }+\text{ }2b\left( 1 \right)\text{ }+\text{ }c\text{ }=\text{ }0 \\
1\text{ }+\text{ }1\text{ }+\text{ }2a\text{ }+\text{ }2b\text{ }+\text{ }c\text{ }=\text{ }0 \\
2a\text{ }+\text{ }2b\text{ }+\text{ }c\text{ }+\text{ }2\text{ }=\text{ }0\ldots ..\text{ }\left( 4 \right) \\
\end{array}\]
Upon simplifying equations (2), (3), (4) we get
a = -13/2, b = -5/2, c = 16
Now by substituting the values of a, b, c in equation (1), we get
\[{{x}^{2}}~+\text{ }{{y}^{2}}~+\text{ }2\text{ }\left( -13/2 \right)x\text{ }+\text{ }2\text{ }\left( -5/2 \right)y\text{ }+\text{ }16\text{ }=\text{ }0{{x}^{2}}~+\text{ }{{y}^{2}}~\text{ }13x\text{ }\text{ }5y\text{ }+\text{ }16\text{ }=\text{ }0\]
∴ The equation of the circle is x2 + y2 – 13x – 5y + 16 = 0