How do you write x^2+y^2+8x+7=0  in standard form?

Jul 6, 2018

${\left(x + 4\right)}^{2} + {y}^{2} = 9$
circle with center $\left(- 4 , 0\right) , \text{ } r = 3$

Explanation:

Given: ${x}^{2} + {y}^{2} + 8 x + 7 = 0$

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:

$\left({x}^{2} + 8 x\right) + {y}^{2} = - 7$

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

We do this because ${\left(x + 4\right)}^{2} = {x}^{2} + 8 x + 16$. 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.

${\left(x + 4\right)}^{2} + {y}^{2} = - 7 + 16$

${\left(x + 4\right)}^{2} + {y}^{2} = 9$

This is the standard form of a circle: ${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$

where the center is $\left(h , k\right)$ and the radius $= r$