The circle passing through the intersection of the circles,
x2 + y2 -- 6x = 0 and x2 + y2 -- 4y = 0, having its centre on
the line, 2x -- 3y + 12 = 0, also passes through the point :
Solutions · 1
Official solution
Let S be the circle passing through point of intersection of S1 \& S2
\[ \therefore \] S = S1 + \[\lambda \]S2 = 0
\[ \Rightarrow \] \[S:({x^2} + {y^2} - 6x) + \lambda ({x^2} + {y^2} - 4y) = 0\]
\[ \Rightarrow \] \[S:{x^2} + {y^2} - \left( {{6 \over {1 + \lambda }}} \right)x - \left( {{{4\lambda } \over {1 + \lambda }}} \right)y = 0\] ....(1)
Centre \[\left( {{3 \over {1 + \lambda }},{{2\lambda } \over {1 + \lambda }}} \right)\] lies on
\[2x - 3y + 12 = 0 \Rightarrow \lambda = - 3\]
put in \[(1) \Rightarrow S:{x^2} + {y^2} + 3x - 6y = 0\]
Now check options point (−3,6)
lies on S.