In the figure below, semicircles with centers at $A$ and $B$ and with radii 2 and 1, respectively, are drawn in the interior of, and sharing bases with, a semicircle with diameter $JK$. The two smaller semicircles are externally tangent to each other and internally tangent to the largest semicircle. A circle centered at $P$ is drawn externally tangent to the two smaller semicircles and internally tangent to the largest semicircle. What is the radius of the circle centered at $P$?
[asy] size(5cm); draw(arc((0,0),3,0,180)); draw(arc((2,0),1,0,180)); draw(arc((-1,0),2,0,180)); draw((-3,0)--(3,0)); pair P = (-1,0)+(2+6/7)*dir(36.86989); draw(circle(P,6/7)); dot((-1,0)); dot((2,0)); dot(P); [/asy]
$\textbf{(A)}\ \frac{3}{4} \qquad \textbf{(B)}\ \frac{6}{7} \qquad\textbf{(C)}\ \frac{\sqrt{3}}{2} \qquad\textbf{(D)}\ \frac{5}{8}\sqrt{2} \qquad\textbf{(E)}\ \frac{11}{12}$
Connect the centers of the tangent circles! (call the center of the large circle $C$)

Notice that we don't even need the circles anymore; thus, draw triangle $\Delta ABP$ with cevian $PC$:

and use [Stewart's Theorem](https://artofproblemsolving.com/wiki/index.php/Stewart%27s_Theorem):
\[AB \cdot AC \cdot BC + AB \cdot {CP}^2 = AC \cdot {BP}^2 + BC \cdot {AP}^2\]
From what we learned from the tangent circles, we have $AB = 3$, $AC = 1$, $BC = 2$, $AP = 2 + r$, $BP = 1 + r$, and $CP = 3 - r$, where $r$ is the radius of the circle centered at $P$ that we seek.
Thus:
\[3 \cdot 1 \cdot 2 + 3 {\left(3-r\right)}^2 = 1 {\left(1+r\right)}^2 + 2 {\left(2+r\right)}^2\]
\[6 + 3\left(9 - 6r + r^2\right) = \left(1 + 2r + r^2\right) + 2\left(4 + 4r + r^2\right)\]
\[33 - 18r + 3r^2 = 9 + 10r + 3r^2\]
\[28r = 24\]
\[r = \boxed{\frac{6}{7}} \to \boxed{\textbf{(B)}}\]