# What is the circle of the equation if the center is (-6,2) passes through the point (-5,3)? Thank you for the answer.

Jun 20, 2018

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

#### Explanation:

$\text{the equation of a circle in standard form is}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{{\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{where "(a,b)" are the coordinates of the centre and r}$
$\text{is the radius}$

$\text{here } \left(a , b\right) = \left(- 6 , 2\right)$

$\text{the radius is the distance from the centre to a point on}$
$\text{the circle}$

$\text{calculate r using the "color(blue)"distance formula}$

•color(white)(x)r=sqrt((x_2-x_1)^2+(y_2-y_1)^2)

$\text{let "(x_1,y_1)=(-6,2)" and } \left({x}_{2} , {y}_{2}\right) = \left(- 5 , 3\right)$

$r = \sqrt{{\left(- 5 + 6\right)}^{2} + {\left(3 - 2\right)}^{2}} = \sqrt{1 + 1} = \sqrt{2}$

$\text{substitute these values into the equation}$

${\left(x - \left(- 6\right)\right)}^{2} + {\left(y - 2\right)}^{2} = {\left(\sqrt{2}\right)}^{2}$

${\left(x + 6\right)}^{2} + {\left(y - 2\right)}^{2} = 2 \leftarrow \textcolor{red}{\text{equation of circle}}$

Jun 20, 2018

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

#### Explanation:

The radius is the distance from the center to any point on the circumference.

Given the center $\left(- 6 , 2\right)$ and the point $\left(- 5 , 3\right)$
The distance between the points parallel to the x-axis is $1$ (the absolute value of the difference between the x-coordinate values), and
the distance between the points parallel to the y-axis is also $1$ (the absolute value of the difference between the y-coordinate values).

By the Pythagorean Theorem this distance (the radius, remember) is $r = \sqrt{{1}^{2} + {1}^{2}} = \sqrt{2}$

The general equation for a circle with center $\left(a , b\right)$ and radius $r$ is
$\textcolor{w h i t e}{\text{XXX}} {\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$

So with a center of $\left(- 6 , 2\right)$ and a radius of $\sqrt{2}$,
we have
$\textcolor{w h i t e}{\text{XXX}} {\left(x + 6\right)}^{2} + {\left(y - 2\right)}^{2} = 2$