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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 $\overline{AD}$?

Rollingblade Dec 1, 2018

#1**+2 **

Very clever, analytic geometry in disguise.

Assuming *D* is on the origin (*D* can be anywhere, the origin is just easier to work with):

Circle *A* has a radius of 4 and center of \((0,4)\), therefore its equation is:

\(x^2+(y-4)^2=16\)

Circle *M* has a radius of 2 and center of \((2,0)\), therefore its equation is:

\((x-2)^2+y^2=4\)

Solving the systems, we get:

\((\frac85, \frac{16}{5}); (0,0)\)

Since *D *is on \((0,0)\), *P *is obviously on \((\frac85, \frac{16}{5})\).

Therefore, the distance is \(\boxed{\frac{16}{5}}\)

I hope this helped,

Gavin.

GYanggg Dec 1, 2018

#2**+1 **

Let A = (0,0) B = (0, 4) C = (4,4) D = (4,0)

Let M = (4, 2)

Let the circle centered at A have the equation

x^2 + y^2 = 16 (1)

Let the circle centered at M have the equation

(x - 4)^2 + ( y - 2)^2 = 4

x^2 - 8x + 16 + y^2 - 4y + 4 = 4

x^2 - 8x + 16 + y^2 - 4y = 0

x^2 + y^2 - 8x - 4y = 0 (2)

Subbing (1) into (2) we can find the y coordinate of P....this will be the distance from P to AD

16 - 8x - 4y = 0

8x = 16 - 4y

2x = 8 - y

x = [ 8 - y] / 2 (3)

Sub (3) into (1)

[(8 - y)/2]^2 + y^2 = 16 simplify

[y^2 - 16y + 64] / 4 + y^2 = 16

y^2 - 16y + 64 + 4y^2 = 64

5y^2 - 16y = 0

y ( 5y - 16) = 0 set each factor to 0 and solve for y

y = 0 [ reject] 5y - 16 = 0

5y = 16

y = 16/5

This is the y coordinate of P.....so..it is 16/5 units from AD

CPhill Dec 1, 2018