# How do you find the center and the radius of the circle whose equation is (x+9)^2 +(y-12)^2=62?

Feb 7, 2016

Radius : $\setminus \sqrt{62}$ units;
Centre : $\setminus \vec{{r}_{0}} = \left(- 9 , + 12\right)$ units,

#### Explanation:

Consider a coordinate system $\setminus \vec{r '} = \left(x ' , y '\right)$ with its origin at the centre of the circle. In this coordinate system the equation of the circle is $x {'}^{2} + y {'}^{2} = {a}^{2}$.

Suppose if the centre of the circle is located at the point $\setminus \vec{{r}_{0}} = \left({x}_{0} , {y}_{0}\right)$ in the original coordinate system.

A point on the circle has a coordinate $\setminus \vec{{r}_{}} = \left(x , y\right)$ in the original coordinate and $\setminus \vec{r '} = \left(x ' , y '\right)$ in the circle centric coordinate. $\setminus \vec{{r}_{}}$ and $\setminus \vec{r '}$ are related by $\setminus \vec{r '} = \setminus \vec{{r}_{}} - \setminus \vec{{r}_{0}}$.

$\setminus \vec{r '} = \setminus \vec{{r}_{}} - \setminus \vec{{r}_{0}}$
$\left(x ' , y '\right) = \left(x , y\right) - \left({x}_{0} , {y}_{0}\right) = \left(x - {x}_{0} , y - {y}_{0}\right)$

$x {'}^{2} + y {'}^{2} = {a}^{2}$
${\left(x - {x}_{0}\right)}^{2} + {\left(y - {y}_{0}\right)}^{2} = {a}^{2}$

Comparing this to the given equation ${\left(x + 9\right)}^{2} + {\left(y - 12\right)}^{2} = 62$, we find that ${x}_{0} = - 9$: $\setminus \quad$ ${y}_{0} = + 12$ and $a = \setminus \sqrt{64}$.

So the position vector of the circle centre in the original coordinate system is $\setminus \vec{{r}_{0}} = \left(- 9 , + 12\right)$. It is at a distance of ${r}_{0} = \setminus \sqrt{{\left(- 9\right)}^{2} + {12}^{2}} = 15$ units at an angle $a \tan \left(\setminus \frac{12}{- 9}\right) = {126.869876}^{o}$ counter-clockwise to the X axis.