Question

Find the equation of circle passing through the points of intersection of $x^2+y^2=6$ and $x^2+y^2-6x+8=0$ and the point $(1,1)$

Answer 1

$S_1=x^2+y^2-6=0$

$S_2=x^2+y^2-6x+8=0$

$S_1+k(S_2-S_1)=0$

$S_2-S_1=-6x+14$

$x^2+y^2-6+k(-6x+14)=0$

$(1,1)$

$1^2+1^2-6+k(-6(1)+14)=0$

$-4+8k=0$

$\therefore k=\frac{1}{2}$

$\Rightarrow \left(2\right)(x^2+y^2-6)+\left(1\right)(-6x+14)=0$

$2x^2+2y^2-12-6x+14=0$

$2x^2+2y^2-6x+2=0$

$\therefore x^2+y^2-3x+1=0$ is the required equation.