Question 206555
How do you find the standard form of the equation of a circle, with a center (3,-1) through the point (-5,1)?
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Find the distance to the point, that's the radius.
r^2 = diffy^2 + diffx^2
r^2 = 2^2 + 8^2
r^2 = 68
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Then it's (x-h)^2 + (y-k)^2 = r^2 (h,k) is the center
(x-3)^2 + (y+1)^2 = 68
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It can be expanded and rearranged.