12. A circle touches the line $2 x + 3 y + 1 = 0$ at the point $( 1 , - 1 )$ and is orthogonal to the circle which has the line segment having end points $( 0 , - 1 )$ and $( - 2,3 )$ as the diameter.
Sol. Let the circle with tangent $2 \mathrm { x } + 3 \mathrm { y } + 1 = 0$ at $( 1 , - 1 )$ be $( x - 1 ) ^ { 2 } + ( y + 1 ) ^ { 2 } + \lambda ( 2 x + 3 y + 1 ) = 0$ or $x ^ { 2 } + y ^ { 2 } + x ( 2 \lambda - 2 ) + y ( 3 \lambda + 2 ) + 2 + \lambda = 0$. It is orthogonal to $\mathrm { x } ( \mathrm { x } + 2 ) + ( \mathrm { y } + 1 ) ( \mathrm { y } - 3 ) = 0$ Or $\mathrm { x } ^ { 2 } + \mathrm { y } ^ { 2 } + 2 \mathrm { x } - 2 \mathrm { y } - 3 = 0$ so that $\frac { 2 ( 2 \lambda - 2 ) } { 2 } \cdot \left( \frac { 2 } { 2 } \right) + \frac { 2 ( 3 \lambda + 2 ) } { 2 } \left( \frac { - 2 } { 2 } \right) = 2 + \lambda - 3 \Rightarrow \lambda = - \frac { 3 } { 2 }$. Hence the required circle is $2 x ^ { 2 } + 2 y ^ { 2 } - 10 x - 5 y + 1 = 0$.