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

• A. $(-1,3)$
• B. $(1,-3)$
• C. $(-3,6)$
• D. $(-3,1)$

Answer 1

The correct option is C $(-3,6)$

Given: $S_1:x^2+y^2-6x=0$
$S_2:x^2+y^2-4y=0$
Now,
$S_1+\lambda (S_1-S_2)=0$
$\Rightarrow x^2+y^2-6x+\lambda (4y-6x)=0$
$\Rightarrow x^2+y^2-6x(1+\lambda )+4\lambda y=0$
Centre $\equiv \left(3\right(1+\lambda ),-2\lambda )$
Since centre lies on the line, $2x-3y+12=0$
$\therefore 6+6\lambda +6\lambda +12=0$
$12\lambda =-18$
$\Rightarrow \lambda =-\frac{3}{2}$
$\therefore x^2$+ $y^2+3x-6y=0$
Check options
$(-3,6)$ is the point