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Write the equation of a circle, with center of (12,-3)

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and containing the point (-12,7)??? Please help!!?
asked Oct 30, 2014 in PRECALCULUS by anonymous

1 Answer

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The standard form of circle equation is (x-h)² + (y-k)² = r².

Center = (h, k) = (12, -3) and the point = (-12, 7).

Radius = The distance between the two points (x1, y1) = (12, -3) and (x2, y2) = (-12, 7).

r = √[(x2 - x1)2 + (y2 - y1)2] = √[(-12 - 12)2 + (7 + 3)2] = √[(24)2 + (10)2] = √(576 + 100) = √(676) = 26.

The circle equation is (x-12)² + (y+3)² = (26)².

answered Oct 31, 2014 by casacop Expert

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