# How do you find the center and radius for  x^2 + y^2 +14x = 0?

Oct 25, 2016

#### Explanation:

The standard form for the equation of a circle is:

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$

Substitute ${\left(y - 0\right)}^{2}$ for ${y}^{2}$

${x}^{2} + 14 x + {\left(y - 0\right)}^{2} = 0$

Using the pattern ${\left(x - h\right)}^{2} = {x}^{2} - 2 h x + {h}^{2}$, we observe that we must begin the process of completing the square by adding ${h}^{2}$ to both sides:

${x}^{2} + 14 x + {h}^{2} + {\left(y - 0\right)}^{2} = {h}^{2}$

For this circle, the ${h}^{2}$ term is, also the ${r}^{2}$ term, because the we will soon find that the center is offset to the left the same distance as the radius.

We can find the value $h$ by setting the right side of the pattern equal to the first 3 terms in the equation:

${x}^{2} - 2 h x + {h}^{2} = {x}^{2} + 14 x + {h}^{2}$

The square terms cancel:

$- 2 h x = 14 x$

#h = -7 and h^2 = 49

Substitute ${\left(x - - 7\right)}^{2}$ for the corresponding terms on the left and

${\left(x - - 7\right)}^{2} + {\left(y - 0\right)}^{2} = 49$

The radius should be represented as a positive number squared.

${\left(x - - 7\right)}^{2} + {\left(y - 0\right)}^{2} = {7}^{2}$

The center is found by observation $\left(- 7 , 2\right)$