# How do you find the center and radius for x^2 + y^2 -12y + 25= 0?

Aug 1, 2016

Centre is thus $\left(0 , 6\right)$ and radius is $\sqrt{11}$

#### Explanation:

We want to complete the squares. As there is no x term of degree one we leave it as is and focus on the y.

${x}^{2} + {y}^{2} - 12 y + 25 = 0$

Subtract 25 from both sides:

${x}^{2} + {y}^{2} - 12 y = - 25$

Square half the coefficient of $y$ and add to both sides:

${x}^{2} + {y}^{2} - 12 y + {\left(- 6\right)}^{2} = - 25 + {\left(- 6\right)}^{2}$

Recognise that because the constant term is half the coefficient we can convert the left hand side to squared form and simplify the right hand side.

${x}^{2} + {\left(y - 6\right)}^{2} = 11$

This is now in the standard form

${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$ for circle centre $\left(a , b\right)$ and radius $r$.

Centre is thus $\left(0 , 6\right)$ and radius is $\sqrt{11}$