Two circles $C_1$ and $C_2$ of radii $10$ cm and $8$ cm are respectively are tangent to each other internally at a point $A$. $AD$ is the diameter of $C_1$ and $P$ and $M$ are points on $C_1$,$C_2$ respectively such that $PM$ is tangent to $C_2$ . If $PM = \sqrt{20}$ and $\angle PAD = x^\circ$, find $x$.
What I Tried: Here is a picture :-
I absolutely have no good ideas to start this. Let $O$ and $K$ be the centers of the $2$ circles. The only thing which I could do was to find $PO$ using Pythagoras Theorem, which will come out to be $\sqrt{84}$.
I also did, for example, checked this in Geogebra and I found $\angle PAD$ to be $60^\circ$, so I suppose that is the answer, but why is it coming so? I didn't find any ideas using Geogenra either.
Can anyone help me? Thank You.
Best Answer
$OK = 10 - 8 = 2$, $KP = 10, OP = \sqrt{84}$
Using cosine law,
$84 = 2^2 + 10^2 - 40 \cos \angle AKP \implies cos \angle AKP = \frac{1}{2}$
$\angle AKP = 60^0$
$\angle PAD = \frac{1}{2}(180^0 - 60^0) = 60^0$
EDIT: Just using Pythagoras,
Drop perp from $P$ to $AK$ which meets $AK$ at say, $X$ and $KX = x$.
Then in $\triangle POX$ and $\triangle PKX$,
$PX^2 = 10^2 - x^2$
$PX^2 = (\sqrt{84})^2 - (x-2)^2$
Equating both, $x = 5 = \frac{AK}{2}$.
Which means perpendicular from $P$ to $AK$ meets it at midpoint, hence $AP = PK = KA = 10$. So, $\triangle APK$ is equilateral.