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 :

Let S be the circle passing through point of intersection of S₁ &amp; S₂<br><br>$\therefore$ S = S₁ + $\lambda$S₂ = 0<br><br>$\Rightarrow$ $S:({x^2} + {y^2} - 6x) + \lambda ({x^2} + {y^2} - 4y) = 0$<br><br>$\Rightarrow$ $$S:{x^2} + {y^2} - \left( {{6 \over {1 + \lambda }}} \right)x - \left( {{{4\lambda } \over {1 + \lambda }}} \right)y = 0$$ ....(1)<br><br>Centre $\left( {{3 \over {1 + \lambda }},{{2\lambda } \over {1 + \lambda }}} \right)$ lies on<br><br>$2x - 3y + 12 = 0 \Rightarrow \lambda = - 3$<br><br>put in $(1) \Rightarrow S:{x^2} + {y^2} + 3x - 6y = 0$<br><br>Now check options point $( - 3,6)$<br><br>lies on S.