# How do you find the center and radius of  (x+3)^2 +y^2 = 37?

Feb 3, 2016

I found: center at $\left(- 3 , 0\right)$ and radius $6.1$.

#### Explanation:

We can compare our equation with the general equation of a circle with center at $\left(h , k\right)$ and radius $r$:
${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$
giving us:
$h = - 3$
$k = 0$
$r = \sqrt{37} \approx 6.1$

Feb 3, 2016

centre (-3 , 0 ) , radius = $\sqrt{37}$

#### Explanation:

the standard form of the equation of a circle is

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

where centre = (a , b ) and r = radius.

the equation here is in this form and so values of a , b and r
can be written down.

here a = -3 , b = 0 and ${r}^{2} = 37 \Rightarrow r = \sqrt{37}$

hence centre = (-3 , 0 )