How do you find the center and radius of the circle given x^2+y^2+6y=-50-14x?

Apr 18, 2018

Centre (-7,-3) radius $\sqrt{8}$

Explanation:

${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$ is the standard form for the equation of a circle with $\left(a , b\right)$ the centre and radius $r$

Rearrange to this form:
${x}^{2} + 14 x + {y}^{2} + 6 y = - 50$

Complete the square:
$\left[{\left(x + 7\right)}^{2} - 49\right] + \left[{\left(y + 3\right)}^{2} - 9\right] = - 50$

${\left(x + 7\right)}^{2} + {\left(y + 3\right)}^{2} = - 50 + 58$

Collect terms and tidy up
${\left(x + 7\right)}^{2} + {\left(y + 3\right)}^{2} = 8$