Centre of the first circle is $(-g,-f)$ and its radius

$C T=r_1=\sqrt{g^2+f^2-c}$

The centre of the second circle is also $(- g ,- f )$ but its radius is

$C P=r_2=\sqrt{g^2+f^2-c \sin ^2 \alpha-\left(g^2-f^2\right) \cos ^2 \alpha}$

or $I _2=\sqrt{ g ^2+ f ^2- c } \cdot \sin \alpha$

$=r_1 \sin \alpha \ldots(1)$

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

$\sin \theta=\frac{r_2}{r_1}=\frac{r_1 \sin \alpha}{r_1}=\sin \alpha$

$\therefore \theta=\alpha \therefore \angle P T A=2 \theta=2 \alpha$