A circle passes through the three vertices of an isosceles triangle that has two sides of length $3$ and a base of length $2$. What is the area of this circle?
$\textbf{(A) } 2\pi \qquad\textbf{(B) } \frac{5}{2}\pi \qquad\textbf{(C) } \frac{81}{32}\pi \qquad\textbf{(D) } 3\pi \qquad\textbf{(E) } \frac{7}{2}\pi$
Solution 1
Let $\triangle ABC$ have vertex $A$ and center $O$, with foot of altitude from $A$ intersecting $BC$ at $D$. 


Then by the Pythagorean Theorem (with radius $r$ and height $OD = h$) on $\triangle OBD$ and $\triangle ABD$
\begin{align*} h^2 + 1 & = r^2 \\ (h + r)^2 + 1 & = 9 \end{align*}
Substituting and solving gives $r = \frac {9}{4\sqrt {2}}$. Then the area of the circle is $r^2 \pi = \left(\frac {9}{4\sqrt {2}}\right)^2 \pi = \boxed{\mathrm{(C) \ } \frac {81}{32} \pi}$.

Solution 2
By $A = \frac {1}{2}Bh = \frac {abc}{4R}$ (or we could use $s = 4$ and [Heron's formula](https://artofproblemsolving.com/wiki/index.php/Heron%27s_formula)),
\[R = \frac {abc}{2Bh} = \frac {3 \cdot 3 \cdot 2}{2(2)(2\sqrt {2})} = \frac {9}{4\sqrt {2}}\]
and the answer is $R^2 \pi = \boxed{\mathrm{(C) \ } \frac {81}{32} \pi}$
Alternatively, by the Extended [Law of Sines](https://artofproblemsolving.com/wiki/index.php/Law_of_Sines),
\[2R = \frac {AC}{\sin \angle ABC} = \frac {3}{\frac {2\sqrt {2}}{3}} \Longrightarrow R = \frac {9}{4\sqrt {2}}\]
Answer follows as above.

Solution 3
Extend segment $AD$ to $R$ on Circle $O$.


By the Pythagorean Theorem \[AD^2 = 3^2 - 1^2\] \[AD = 2\sqrt{2}\]
$\triangle ADC$ is similar to $\triangle ACR$, so \[\frac {2\sqrt{2}}{3} = \frac {3}{2r}\] which gives us \[2r = \frac {9}{2\sqrt{2}} = \frac {9\sqrt{2}}{4}\] therefore \[r = \frac {9\sqrt{2}}{8}\]
The area of the circle is therefore $\pi r^2 = \left(\frac {9\sqrt{2}}{8}\right)^2 \pi = \boxed{\mathrm{(C) \ } \frac {81}{32} \pi}$

Solution 4
First, we extend $AD$ to intersect the circle at $E.$


By the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem), we know that 
\[1^2+AD^2=3^2\implies AD=2\sqrt{2}.\]
We also know that, from the [Power of a Point Theorem](https://artofproblemsolving.com/wiki/index.php/Power_of_a_Point_Theorem), \[AD\cdot DE=BD\cdot DC.\] 
We can substitute the values we know to get
\[2\sqrt{2}\cdot DE=1\]
We can simplify this to get that 
\[DE=\dfrac{2\sqrt{2}}{8}.\]
We add $AD$ and $DE$ together to get the length of the diameter, and then we can find the area.
\[AE=AD+DE=2\sqrt{2}+\dfrac{2\sqrt{2}}{8}=\dfrac{9\sqrt{2}}{4}.\]
Therefore, the radius is $\dfrac{9\sqrt{2}}{8}$, so the area is 
\[\left(\dfrac{9\sqrt{2}}{8}\right)^2\pi=\boxed{(\text{C})\dfrac{81}{32}\pi.}\]

Solution 5
Another possible solution is to plot the circle and triangle on a graph with the circle having center $(0,0)$. 
Let the radius of the circle = $r$. 
Let the distance between the origin and the base of triangle = $a$.
\[1 + a^2 = r^2\]
\[r + a = 2\sqrt{2}\]
\[a = 2\sqrt{2} - r\]
\[9 - 4r\sqrt{2} = 0\]
\[r = \frac{9\sqrt{2}}{8}\]
\[\pi r^2 = \boxed{(\text{C})\frac{81}{32}\pi}\]

Solution 6
We can also use LOC to solve this problem. If O is the center of the circle, angle $OBC$ is double angle $BAC$. From law of cosines, we have that $\cos(\angle BAC) = \frac{7}{9}$. We now apply law of cosines on triangle $OBC$ to get that $4 = r^2+r^2-2r^2\cos(2 \angle BAC)$ but we know that $2 \cos (\angle BAC) = 2 \cdot (\frac{7}{9})^2 - 1 = \frac{17}{81}$. Plugging this in and simplifying, we get $r = \frac{9}{8} \cdot \sqrt{2}$ and thus the area of the circle is equal to $\frac{81}{32} \pi$.