SOLUTION: What is the equation in standard form for the circle: x^2+y^2-4x+8y+4=0

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Question 298099: What is the equation in standard form for the circle: x^2+y^2-4x+8y+4=0
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Rewrite in standard form:
x%5E2%2By%5E2-4x%2B8y%2B4+=+0 Group the x-terms and the y-terms as shown:
%28x%5E2-4x%29%2B%28y%5E2%2B8y%29%2B4+=+0 Subtract 4 from both sides.
x%5E2-4x%29%2B%28y%5E2%2B8y%29+=+-4 Now complete the square in x and y by adding the square of half the x-term coefficient and the square of half the y-term coefficient to both sides of the equation.
Simplify both sides of the equation.
%28x%5E2-4x%2B4%29%2B%28y%5E2%2B8y%2B16%29+=+-4%2B4%2B16 Now factor the left side and simplify the right side.
%28x-2%29%5E2%2B%28y%2B4%29%5E2+=+16 Finally...
highlight%28%28x-2%29%5E2%2B%28y%2B4%29%5E2+=+4%5E2%29 Compare with standard form...
%28x%2Bh%29%5E2%2B%28y%2Bk%29%5E2+=+r%5E2 and you can see that h+=+-2,k+=+4, and r+=+4 and this represents a circle with center at (-2, 4) and radius of 4.