SOLUTION: what is the equation of the circle that has its center at (3,-1) and passes through the point (-2,2)

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Question 374779: what is the equation of the circle that has its center at (3,-1) and passes through the point (-2,2)
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
%28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2 Start with the general equation of a circle.


%28x-3%29%5E2%2B%28y--1%29%5E2=r%5E2 Plug in h=3 and k=-1 (since the center is the point (h,k) ).


%28-2-3%29%5E2%2B%282--1%29%5E2=r%5E2 Plug in x=-2 and y=2 (this is the point that lies on the circle, which is in the form (x,y) ).


%28-5%29%5E2%2B%283%29%5E2=r%5E2 Combine like terms.


25%2B9=r%5E2 Square each term.


34=r%5E2 Add.


So because h=3, k=-1, and r%5E2=34, this means that the equation of the circle with center (3,-1) that goes through the point (-2,2) is


%28x-3%29%5E2%2B%28y%2B1%29%5E2=34.


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