Question

From any point on the circle $x^2+y^2+2gx+2fy+c=0$ tangents are drawn to the circle $x^2+y^2+2gx+2fy+c{\sin }^2\alpha +(g^2+f^2){\cos }^2\alpha =0$; prove that the angle between them is $2\alpha$.

Answer 1

From the figure, we see that, $r_1^2=r_2^2{\sin }^2\theta$, where $2\theta$ is the angle between the $2$ tangents

$r_2^2=g^2+f^2-c$

$r_1^2=g^2+f^2-c{\sin }^2\alpha -(g^2+f^2){\cos }^2\alpha$

$\Rightarrow r_1^2=(g^2+f^2-c){\sin }^2\alpha =r_2^2{\sin }^2\alpha$

Which gives $\theta =\alpha$

$\therefore$ the angle between the two tangents equals $2\alpha$