# Where is the center of the circle given by the equation (x - 3)^2 + (y + 2)^2 = 9?

May 25, 2018

the centre is$\text{ } \left(3 , - 2\right)$

#### Explanation:

an eqn of a circle is given by

${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2} - - \left(1\right)$

where the centre is $\left(a , b\right)$

and the radius is $r$

for

${\left(x - 3\right)}^{2} + {\left(y + 2\right)}^{2} = 9$

comparing with $\left(1\right)$

the centre is$\left(3 , - 2\right)$

as an extra we see the radius is

$r = \sqrt{9} = 3$