The circle passing through the intersection of the circles x^2+y^2-6x=0 and x^2+y^2-4y=0, …

Question

The circle passing through the intersection of the circles $x^2+y^2-6x=0$ and $x^2+y^2-4y=0$, having its centre on the line $2x-3y+12=0$, also passes through the point:

Accepted Answer

Family: $(x^2+y^2-6x)+\lambda(x^2+y^2-4y)=0$, centre $\left(\dfrac3{1+\lambda},\dfrac{2\lambda}{1+\lambda}\right)$.

On $2x-3y+12=0$: $\dfrac{6-6\lambda}{1+\lambda}+12=0\Rightarrow6-6\lambda+12+12\lambda=0\Rightarrow\lambda=-3$.

Circle: $x^2+y^2+3x-6y=0$. Checking $(-3,6)$: $9+36-9-36=0$ .

Solutions · 1