Show that length of tangent from any point on the circle$ x^2 + y^2 + 2gx + 2fy +c=0 $ to the circle $x^2 + y^2 + 2gx + 2fy +c_1 =0 $ is $\sqrt {c}-c_1$
$\endgroup$ 12 Answers
$\begingroup$Let $C_1:x^2+y^2+2gx+2fy+c=0$ and $C_2=x^2+y^2+2gx+2fy+c_1=0$. Now note that $C_1=(x+g)^2+(y+f)^2=f^2+g^2-c$ and $C_2=(x+g)^2+(y+f)^2=f^2+g^2-c_1$. Now assuming that $c\ge c_1$, $C_1$ is contained in $C_2$ and the length of the tangent from any point on $C_1$ to $C_2$ is $r_2^2-r_1^2=\sqrt{c-c_1}$, using the fact that tangent of the circle is perpendicular to the radius at the point of tangency.
$\endgroup$ $\begingroup$The concentric circles imply a fixed length of tangent from any location on the $c_1$ circle.
the following diagram illustrates the solution.
