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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 :
Rearranging and adding like terms yields:
`-16sqrt(36-x^2)=10x-109` Square both sides:
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`
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