# How do you find the center and radius of the circle given by the equation x^2+y^2-8 x- 6 y +21=0?

Jan 8, 2016

Rearrange into the form:

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

to identify the centre $\left(4 , 3\right)$ and radius $2$.

#### Explanation:

Complete the squares for $x$ and $y$...

$0 = {x}^{2} + {y}^{2} - 8 x - 6 y + 21$

$= \left({x}^{2} - 8 x + 16\right) + \left({y}^{2} - 6 y + 9\right) - 4$

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

Add ${2}^{2}$ to both ends and transpose to get:

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

This is in the form:

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

which is the equation of a circle with centre $\left(h , k\right) = \left(4 , 3\right)$ and radius $r = 2$

Notice that I picked out the constant value $16$ to complete the square ${\left(x - 4\right)}^{2} = {x}^{2} - 8 x + 16$ and $9$ to complete the square ${\left(y - 3\right)}^{2} = {y}^{2} - 6 y + 9$.