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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 $\overline{AD}$?

 Dec 1, 2018
 #1
avatar+987 
+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 is on \((0,0)\)is obviously on \((\frac85, \frac{16}{5})\)

 

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

 

I hope this helped,

 

Gavin.

 Dec 1, 2018
 #2
avatar+101796 
+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

 

 

cool cool cool

 Dec 1, 2018

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