Square $ABCD$ has sides of length $4$, and $M$ is the midpoint of $\overline{CD}$. A circle with radius $2$ and center $M$ intersects a circle with radius $4$ and center $A$ at points $P$ and $D$. What is the distance from $P$ to $\overline{AD}$?

$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ \frac {16}{5} \qquad \textbf{(C)}\ \frac {13}{4} \qquad \textbf{(D)}\ 2\sqrt {3} \qquad \textbf{(E)}\ \frac {7}{2}$
Let $D$ be the origin. $A$ is the point $(0,4)$ and $M$ is the point $(2,0)$. We are given the radius of the quarter circle and semicircle as $4$ and $2$, respectively, so their equations, respectively, are:
$x^2 + (y-4)^2 = 4^2$
$(x-2)^2 + y^2 = 2^2$
Subtract the second equation from the first:
$x^2 + (y - 4)^2 - (x - 2)^2 - y^2 = 12$
$4x - 8y + 12 = 12$
$x = 2y.$
Then substitute:
$(2y)^2 + (y - 4)^2 = 16$
$4y^2 + y^2 - 8y + 16 = 16$
$5y^2 - 8y = 0$
$y(5y - 8) = 0.$
Thus $y = 0$ and $y = \frac{8}{5}$ making $x = 0$ and $x = \frac{16}{5}$.
The first value of $0$ is obviously referring to the x-coordinate of the point where the circles intersect at the origin, $D$, so the second value must be referring to the x coordinate of $P$. Since $\overline{AD}$ is the y-axis, the distance to it from $P$ is the same as the x-value of the coordinate of $P$, so the distance from $P$ to $\overline{AD}$ is $\frac{16}{5} \Rightarrow \boxed{B}.$
-mathbeast9
(JK lol)