How do you identify the center and radius of the circle x^2+(y+12)^2=24?

Feb 23, 2017

$\text{centre "(0,-12)" radius } r = 2 \sqrt{6}$

Explanation:

$\text{The standard equation of a circle with centre "(a,b)" and radius " r " is}$

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

$\text{We have } {x}^{2} + {\left(y + 12\right)}^{2} = 24$

$\text{comparing this with the standard eqn.}$

$a = 0 , b = - 12$

${r}^{2} = 24 \implies r = \sqrt{24} = 2 \sqrt{6}$

$\text{centre "(0,-12)" radius } r = 2 \sqrt{6}$

Feb 23, 2017

$\text{centre "=(0,-12)," radius } = 2 \sqrt{6}$

Explanation:

The standard form of the $\textcolor{b l u e}{\text{equation of a circle}}$ is.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{{\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
where (a ,b) are the coordinates of the centre and r, the radius.

${x}^{2} + {\left(y + 12\right)}^{2} = 24 \text{ is in this form}$

by comparison with the standard form equation.

$a = 0 , b = - 12 \text{ and } {r}^{2} = 24$

${r}^{2} = 24 \Rightarrow r = \sqrt{24} = 2 \sqrt{6}$

$\Rightarrow \text{centre "=(a,b)=(0,-12)" and } r = 2 \sqrt{6}$