# Circle 1 is centered at (0,0) and is represented by the equation x^2+y^2=36. Circle 2 is centered at (5,8) with a radius of 4 units. verify algebrically......that the circles overlap.

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Prove that the circles given by `x^2+y^2=36` and the circle centered at (5,8) with radius 4 overlap.

(1) The equation for the circle centered at (5,8) with radius 4 is `(x-5)^2+(y-8)^2=16`

(2) One way to prove the circles intersect is to find the intersection.

(a) Expand the equation for the second circle:

`(x-5)^2+(y-8)^2=16 => x^2-10x+25+y^2-16x+64=16`

(b) From the first circle we have `y^2=36-x^2,y=sqrt(36-x^2)`

(c) Substituting we get :

`x^2-10x+25+36-x^2-16sqrt(36-x^2)+64=16`

Rearranging and adding like terms yields:

`-16sqrt(36-x^2)=10x-109` Square both sides:

`256(36-x^2)=100x^2-2180x+11881`

`356x^2-2180x+2665=0`

So `x=(2180+-sqrt(2180^2-4(356)(2665)))/(2(356))`

`x=(2180+-32sqrt(935))/712=1/178(545+-8sqrt(935))`

Plugging in to either equation to find y we get the points of intersection to be `(1/178(545-8sqrt(935)),1/178(872+5sqrt(935)))" and " (1/178(545+8sqrt(935)),1/178(872-5sqrt(935)))` **The fact that we found the points of intersection proves that the circles overlap.**

(3) Another method is to note that the circle centered at the origin intersects the line joining the centers further from the origin than the other circle -- thus the two circles must overlap.

The line joining the centers is `y=8/5x` . The first circle intersects the line when `x^2+(8/5x)^2=36" or " x=30/sqrt(89)~~3.72` while the second circle intersects the line when `(x-5)^2+(8/5x-8)^2=16 => x=-5/89(4sqrt(89)-89)~~2.88`