Two circles each of which passing through the points $(0,k)$ and $(0,-k)$ and touch the line $y=mx+c$. Then prove that they will touch orthogonally if $c^2=k^2(2+m^2)$.
My attempt: Let the equations of the two circles are $x^2+y^2+2g_1x+2f_1y+d_1=0$ and $x^2+y^2+2g_2x+2f_2y+d_2=0$. Since both are passing through $(0,k)$, $(0,-k)$…then by putting this points we get $f_1=f_2=0$ as $k$ is non-zero. And from there we also get $d_1=d_2=-k^2$. Therefore our new equations are $x^2+y^2+2g_1x-k^2=0$ and $x^2+y^2+2g_2x-k^2=0$. If these two circles are orthogonal then $g_1g_2=-k^2$. Now from here how can I eliminate $g_1$ and $g_2$ to get the required condition. Please help me to solve this.
Best Answer
To find $g_1$ and $g_2$ you must use the tangency condition: substituting $y=mx+c$ in the circle equations, the discriminant of the resulting equations must vanish. This leads to $$ (mc+g)^2-(1+m^2)(c^2-k^2)=0. $$ I wrote simply $g$ because the two equations are identical, so that $g_1$ and $g_2$ are just the two solutions of the above equation. From that equation you can then immediately find $g_1g_2=(m^2+1)k^2-c^2$ and equating that to $-k^2$ you get the required relation.