Question

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

Answer 1

Centre of the first circle is $(-g,-f)$ and its radius
$CT=r_1=\sqrt{g^2+f^2-c}$
The centre of the second circle is also $(-g,-f)$ but its radius is
$CP=r_2=\sqrt{g^2+f^2-c{\sin }^2\alpha -(g^2-f^2){\cos }^2\alpha }$
or $r_2=\sqrt{g^2+f^2-c}.\sin \alpha$
$=r_1\sin \alpha .....\left(1\right)$
Since $\sin \alpha$ is less than $1$ therefore $r_2<r_1$ and as such the second circle is inner circle concentric with outer circle. Now if $\theta$ be the angle $PTC$, then from right angles triangle
$\sin \theta =\frac{r_2}{r_1}=\frac{r_1\sin \alpha }{r_1}=\sin \alpha$
$\therefore \theta =\alpha \therefore \angle PTA=2\theta =2\alpha$.