SOLUTION: How do you find the standard form of the equation of a circle, with a center (3,-1) through the point (-5,1)?

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)?
Found 2 solutions by Alan3354, tientrinh:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
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.
Answer by tientrinh(1) (Show Source): You can put this solution on YOUR website!
Find the distance to the point, that's the radius. By the formula d= √(x2-x1)^2+ (y2-y1)^2

d= √(3+5)^2+ (-1-1)^2
d= √64+4 = √68 (or =r)
r^2=68
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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