Problem:
Square ABCD has sides of length 4 , and M is the midpoint of 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 AD ?
Answer Choices:
A. 3
B. 516
C. 413
D. 23
E. 27 Solution:
Place an xy-coordinate system with origin at D and points C and A on the positive x - and y-axes, respectively. Then the circle centered at M has equation
(x−2)2+y2=4
and the circle centered at A has equation
x2+(y−4)2=16
Solving these equations for the coordinates of P gives x=16/5 and y=8/5, so the answer is 16/5.
OR
We have AP=AD=4 and PM=MD=2, so △ADM is congruent to △APM, and ∠APM is a right angle. Draw PQ and PR perpendicular to AD and CD, respectively. Note that ∠APQ and ∠MPR are both complements of ∠QPM. Thus △APQ is similar to △MPR, and
MRAQ=MPAP=24=2
Let MR=x. Then AQ=2x,PR=QD=4−2x, and PQ=RD=x+2. Therefore
