In the figure, $\overline{AB}$ and $\overline{CD}$ are diameters of the circle with center $O$, $\overline{AB} \perp \overline{CD}$, and chord $\overline{DF}$ intersects $\overline{AB}$ at $E$. If $DE = 6$ and $EF = 2$, then the area of the circle is 

$\mathrm{(A) \ 23 \pi } \qquad \mathrm{(B) \ \frac {47}{2} \pi } \qquad \mathrm{(C) \ 24 \pi } \qquad \mathrm{(D) \ \frac {49}{2} \pi } \qquad \mathrm{(E) \ 25 \pi }$
Solution 1
Let the radius of the circle be $r$ and let $x=\overline{OE}$. 
By the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem), $OD^2+OE^2=DE^2 \Rightarrow r^2+x^2=6^2=36$. 
By [Power of a point](https://artofproblemsolving.com/wiki/index.php/Power_of_a_point), $AE \cdot EB = DE \cdot EF \Rightarrow (r+x)(r-x)=r^2-x^2=6\cdot2=12$.  
Adding these equations yields $2r^2=48 \Rightarrow r^2 = 24$. 
Thus, the area of the circle is $\pi r^2 = 24\pi \Rightarrow C$.
Solution 2
Let the radius of the circle be $r$.
We can see that $\triangle CFD$ has a right angle at $F$ and that $\triangle EOD$ has a right angle at $O$.  
Both triangles also share $\angle ODE$, so $\triangle CFD$ and $\triangle EOD$ are [similar](https://artofproblemsolving.com/wiki/index.php?title=Similar(geometry)&action=edit&redlink=1).
This means that $\frac {DE}{OD}=\frac {CD}{FD}$.
So,  $\frac {6}{r}=\frac {2r}{8}$. Simplifying, $r^2 = 24$.
This means the area is $24\pi \Rightarrow C$.