Question

The centre of the circle $S=0$ lie on the line $2x-2y+9=0$ & $S=0$ cuts orthogonally the circle $x^2+y^2=4$. Show that circle $S=0$ passes through two fixed points & find their coordinates.

Answer 1

$r_2^{+}4=x^2+{\left(\frac{2x+9}{2}\right)}^2$

$x^2+y^2+2gx+2fy+c=0$ ....(1)$x^2+y^2-4=0$ ....(2)

orthogonally

$2g_1{g}_2+2f_1{f}_2=c_1+c_2$

$2g\times 0+2f\times 0=c-4$

$c=4$ ---(iii)$(-g,-f)leg2x-2y+9=0$

$-2g+2f+9=0$

$2g=2=2f+g$ ____ (iv)

$x^2+y^2+(2f+g)x+2+y+4=0$

$(x^2+y^2+9x+4)+2+(x+y)=0$

$x^2+y^2+(2f+g)x+2+y+4=0$

$(x^2+y^2+9x+4)+2f(x+y)=0$

$s=x^2+y^2+9x+4=0$

$L=x+y=0$

$x^2+y^2+9x+4=0$

$2x^2+9x+4=0$

$2x^2+8xfx+4=0$

$2x(x+4)+1(x+4)=0$

$x=-1/2,-4$

$x=-1/2,-4\Rightarrow x+y=0$

$y=+1/2,y=4$

$forx(-1/2,1/2)\&(-4,4)$