Find the area of the circle with extern point Points $A, B, C, D$ are on a circle such that $AB = 10$ and $CD = 7$. If $AB$ and $CD$ are extended past $B$ and $C$, respectively, they meet at $P$ outside the circle. Given that $BP = 8$ and $\angle APD = 60^\circ$, ﬁnd the area of the circle.

I have no idea.. Could someone help? I can find the length of $PC$ by power of a point, and maybe use Ptomely's Theorem, but how can I find the area of the circle?
 A: The first step is to add a point $H$ to the figure, located half way between the points $A$ and $B$. Note that the line segment $MH$ is perpendicular to the line through the points $A$, $H$, $B$ and $P$. So we can apply Pythagoras to the triangle $MHP$, yielding:
$$13^2 + (MH)^2 = (MP)^2$$
Similarly we apply Pythagoras to the smaller triangle $MHB$:
$$5^2 + (MH)^2 = R^2$$
Subtracting the second equation from the first gives an expression for the distance $(MP)$ between the centre $M$ and the point $P$ in terms of the radius $R$: 
$$(MP)^2 = R^2 + 144$$
The second step is to apply the same procedure to the other side of the picture. We first define the point $I$, half way between the points $D$ and $C$. The angle between the line segment $MI$ and the line through $D$, $I$, $C$, $P$ is $90$ degrees. So we can apply Pythagoras to the triangles $MIP$ and $MIC$. In combination with the result already obtained for $(MP)$, we find that $[(PC) + 3.5]^2 - 3.5^2 = 144$ from which it follows that the length $(PC)$ is equal to $9$. 
The third step is to derive a formula for the radius $R$ in terms of the other lengths. To do this we use the final bit of information, namely that the angle $APD$ is equal to $60$ degrees. Take the cosine of this angle, use the fact that angle $APD$ = angle $APM$ + angle $DPM$, and apply the well-known rule for the cosine of a sum of two angles. Substituting all our results so far, we get a somewhat messy equation. After tidying up we obtain:
$$R^4 + 71R^2 - 10512 = 0$$
This is a quadratic equation for $R^2$ which has the solution $R^2 = 73$. Therefore the surface area of the circle equals $\pi * 73$.    
A: The clue given is: Circle and its area depend  only upon the power and included angle between any two secants or tangents that you are free to choose. 
So we can decide to settle for tangents squared or power conveniently.
If power is $T^2$, radius R, half included angle $ \beta$, then
Area = $\pi R^2 = \pi (T \tan \beta)^2 $
In this case $ \beta = 30^{0}, T =\sqrt {PB\cdot PA} = \sqrt {80} = 4 \sqrt 5, $ Area = $ \pi \ 80/3. $

EDIT1:
There is an error about pair of secants' included angle considered. My main attempt however is to find Circle area in terms of its power the included angle of  secants only.. that I believe would be more instructive and elegant.
