# How do you find the coordinates of the center of a circle whose equation is x^2-14x+y^2+6y+54=0?

Jan 28, 2016

Complete squares and rearrange into the standard form of the equation of a circle to find centre $\left(h , k\right) = \left(7 , - 3\right)$ and radius $2$

#### Explanation:

Complete the squares for both $x$ and $y$:

$0 = {x}^{2} - 14 x + {y}^{2} + 6 y + 54$

$= \left({x}^{2} - 14 x + 49\right) + \left({y}^{2} + 6 y + 9\right) + \left(54 - 49 - 9\right)$

$= {\left(x - 7\right)}^{2} + {\left(y + 3\right)}^{2} - 4$

Add $4$ to both ends and slightly re-express to get:

${\left(x - 7\right)}^{2} + {\left(y - \left(- 3\right)\right)}^{2} = {2}^{2}$

This is in the standard form of the equation of a circle:

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

with centre $\left(h , k\right) = \left(7 , - 3\right)$ and radius $r = 2$.

graph{(x^2-14x+y^2+6y+54)((x-7)^2+(y+3)^2-0.01) = 0 [-6.79, 13.21, -6.16, 3.84]}