SOLUTION: Find the center of a circle defined by the equation: x^2+y^2-10x+12y+12=0.

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Question 388025: Find the center of a circle defined by the equation: x^2+y^2-10x+12y+12=0.
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!


Hi

Standard Form of an Equation of a Circle is %28x-h%29%5E2+%2B+%28y-k%29%5E2+=+r%5E2

where Pt(h,k) is the center

x^2+y^2-10x+12y+12=0

completing the squares to put into the Standard Form

x^2 -10x + y^2 +12y + 12 = 0

 (x-5)^2 - 25 + (y+6)^2 - 36 + 12 = 0

  (x-5)^2 + (y+6)^2 -49 = 0

  (x-5)^2 + (y+6)^2 = 49 

Center (5,-6)  radius = 7