What is area of circle given areas of $4$ squares 
I can't find a way to solve this, I know that by using power of the point the other square area is 64cm^2 and that its side length is 8, but for the area of the circle I'm not too sure where to start, can someone help?
 A: 
The area of the triangle $ABC$ is 
$$Area_{ABC} = \frac 12 BX\cdot AC = \frac12 \cdot 5\cdot (4+10) = 35$$
Then, the circumradius of $ABC$ is,
$$R = \frac{AB\cdot BC\cdot CA}{4Area_{ABC}}
=\frac{\sqrt{4^2+5^2}\cdot \sqrt{10^2+5^2}\cdot 14}{4\cdot35}
=\frac{\sqrt{205}}{2}$$
Thus, the area of the circle is
$$Area_{circle} = \pi R^2= \frac{205\pi}4$$
A: Let $X$ be the intersection point.   
Hint: Recall that $4 \times 10 = Pow(X) = R^2 - d^2 $.   

 Let $X = (0,0)$.
 The center of the circle is $( -1.5, -3)$, so we can find $d^2$, so we can find $R^2$

A: 
I added a coordinate system to the picture. 


*

*Line $\overleftrightarrow{OG}$ is the perpendicular bisector of chord $\overline{AC}$.

*Line $\overleftrightarrow{OF}$ is the perpendicular bisector of chord $\overline{BD}$.

*Point $O$, the intersection of lines $\overleftrightarrow{OG}$ and $\overleftrightarrow{OF}$ is the center of the circle.


Find the coordinates of point $O$ and the measures of one of the distances to point $A$, $B$, $C$, or $D$. That will give you the radius of the circle.
