SOLUTION: Find the center of the circle defined by the equation: 4x^2+4y^2+24x-56y+168=0.

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Question 385414: Find the center of the circle defined by the equation: 4x^2+4y^2+24x-56y+168=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        

4x^2+4y^2+24x-56y+168 = 0 

4x^2+ 24x + 4y^2-56y+168 = 0 

x^2 +6x +y^2 - 14y + 42 = 0 Completing both squares

 (x+3)^2 -9 + (y -7)^2 -49 + 42 = 0

 (x+3)^2 (y -7)^2 -16 = 0

 (x+3)^2 (y -7)^2 = 16 

Center is Pt(-3,7)