# How do you find the center and radius of the circle given x^2+y^2-6y-16=0?

Jun 5, 2018

#### Answer:

The process is complete the squares. See below

#### Explanation:

Remember that ${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$ and
${\left(a - b\right)}^{2} = {x}^{2} - 2 a b + {b}^{2}$

Then
${x}^{2} + {y}^{2} - 6 x - 16 = {\left(x - 3\right)}^{2} + {y}^{2} - 16 - 9 = {\left(x - 3\right)}^{2} + {y}^{2} - 25 = 0$

Then we have finally ${\left(x - 3\right)}^{2} + {y}^{2} = 25 = {5}^{2}$

This is the equation of a circle centered in $\left(3 , 0\right)$ and radius 5