Question

The radical axis of the circles $x^2+y^2+2gx+2fy+c=0$ and $2x^2+2y^2+3x+8y+2c=0$ touches the circles $x^2+y^2+2x+2y+1=0$; show that either $g=\frac{3}{4}$ or $f=2$.

Answer 1

The second circle can be reduced to standard form on dividing by $2$ so that its equation is
$S_2=x^2+y^2+\frac{3}{2}x+4y+c=0$.
The radical axis of circles $S_1=0$ and $S_2=0$ is given by $S_1-S_2=0$.
or $2x(g-3/4)+2y(f-2)=0$
or $x(g-3/4)+y(f-2)=0$
It touches the circle $S_3=0$ whose centre is $(-1,-1)$ and radius $1$. The condition of tangency i.e. $p=r$ gives
$\frac{-1.(g-3/4)-(f-2)1}{\sqrt{\left[\right(g-3/4{)}^2+(f-2{)}^2]}}=1$.
Squaring, we get
$(g-3/4{)}^2+(f-2{)}^2+2(g-3/4)(f-2)=(g-3/4{)}^2+(f-2{)}^2$
$\therefore 2(g-3/4)(f-2)=0$.
Hence either $g-3/4=0$ or $f-2=0$
$\therefore$ Either $g=3/4$ or $f=2$.