+235 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 $\overline{AD}$?
+981 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.
+128475 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
