SOLUTION: Find the equation of a circle with center at (3,-1) passing through the point (8,-11).

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Question 502987: Find the equation of a circle with center at (3,-1) passing through the point (8,-11).
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) ).


%288-3%29%5E2%2B%28-11--1%29%5E2=r%5E2 Plug in x=8 and y=-11 (this is the point that lies on the circle, which is in the form (x,y) ).


%285%29%5E2%2B%28-10%29%5E2=r%5E2 Combine like terms.


25%2B100=r%5E2 Square each term.


125=r%5E2 Add.


So because h=3, k=-1, and r%5E2=125, this means that the equation of the circle with center (3,-1) that goes through the point (8,-11) is


%28x-3%29%5E2%2B%28y%2B1%29%5E2=125.