Question 968306: Find the coordinates of the center of a circle containing the points (0,0), (-2,4) and (4,-2)
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Find the coordinates of the center of a circle containing the points (0,0), (-2,4) and (4,-2)
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standard form of equation for a circle: (x-h)^2+(y-k)^2=r^2, (h,k)=coordinates of center, r=radius
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(0,0) (0-h)^2+(0-k)^2=r^2
h^2+k^2=r^2
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(-2,4) (-2-h)^2+(4-k)^2=r^2
4+4h+h^2+16-8k+k^2=r^2
4+4h+h^2+16-8k+k^2=h^2+k^2 (sub (h^2+k^2) for r^2)
4+4h+16-8k=0
4h-8k=-20
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(4,-2) (4-h)^2+(-2-k)^2=r^2
16-8h+h^2+4+4k+k^2=r^2
16-8h+h^2+4+4k+k^2=h^2+k^2 (sub (h^2+k^2) for r^2)
16-8h+4+4k=0
-8h+4k=-20
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4h-8k=-20
-8h+4k=-20
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8h-16k=-40(mult. eq. by 2)
-8h+4k=-20
add:
-12k=-60
k=5
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4h=-20+8k=-20+40
4h=20
h=5
r^2=h^2+k^2=25+25=50
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equation: (x-5)^2+(y-5)^2=50
center:(5, 5)
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