How do you write x2+y2+8x+7=0 in standard form?

1 Answer
marfre
Jul 6, 2018

(x+4)2+y2=9
circle with center (−4,0), r=3

Explanation:

Given: x2+y2+8x+7=0

You must use completing of the square. First group the x-terms together, the y-terms together and put the constants on the right side:

(x2+8x)+y2=−7

To complete the square, half the 8x term constant = 4 and add the square of this constant (4^2 = 16) to the right side.

We do this because (x+4)2=x2+8x+16. When the square is completed there will always be a constant that needs to be added to the other side to keep the equation balanced.

(x+4)2+y2=−7+16

(x+4)2+y2=9

This is the standard form of a circle: (x−h)2+(y−k)2=r2

where the center is (h,k) and the radius =r

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