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 $\overline{AD}$? [asy] pair A,B,C,D,M,P; D=(0,0); C=(10,0); B=(10,10); A=(0,10); M=(5,0); P=(8,4); dot(M); dot(P); draw(A--B--C--D--cycle,linewidth(0.7)); draw((5,5)..D--C..cycle,linewidth(0.7)); draw((7.07,2.93)..B--A--D..cycle,linewidth(0.7)); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$M$",M,S); label("$P$",P,N); [/asy]
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 $\frac{16}{5} \Rightarrow \fbox{\frac {16}{5}}.$ -mathbeast9