# How do find the standard equation of the circle with center (3,6) and passes through (-1,4)?

Aug 25, 2017

${\left(x - 3\right)}^{2} + {\left(y - 6\right)}^{2} = 20$

#### Explanation:

Standard equation of a circle with center $\left(h , k\right)$ and radius $r$ is

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

As center is $\left(3 , 6\right)$ and circle passes through $\left(- 1 , 4\right)$, the radius is distance between $\left(3 , 6\right)$ and $\left(- 1 , 4\right)$ i.e.

$r = \sqrt{{\left(3 - \left(- 1\right)\right)}^{2} + {\left(6 - 4\right)}^{2}} = \sqrt{{4}^{2} + {2}^{2}} = \sqrt{16 + 4} = \sqrt{20}$

Hence standard equation of circle is

${\left(x - 3\right)}^{2} + {\left(y - 6\right)}^{2} = {\left(\sqrt{20}\right)}^{2}$

or ${\left(x - 3\right)}^{2} + {\left(y - 6\right)}^{2} = 20$

graph{((x-3)^2+(y-6)^2-20)((x-3)^2+(y-6)^2-0.02)((x+1)^2+(y-4)^2-0.02)=0 [-8.13, 11.87, 0.64, 10.64]}

Aug 25, 2017

${\left(x - 3\right)}^{2} + {\left(y - 6\right)}^{2} = 20$

#### Explanation:

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

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

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

$\text{the radius is the distance from the centre to the }$
$\text{point } \left(- 1 , 4\right)$

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

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

$\text{with "(x_1,y_1)=(3.6)" and } \left({x}_{2} , {y}_{2}\right) = \left(- 1 , 4\right)$

$\Rightarrow r = \sqrt{{\left(- 1 - 3\right)}^{2} + {\left(4 - 6\right)}^{2}} = \sqrt{20}$

$\text{here "(a,b)=(3,6)" and } {r}^{2} = {\left(\sqrt{20}\right)}^{2} = 20$

$\Rightarrow {\left(x - 3\right)}^{2} + {\left(y - 6\right)}^{2} = 20 \leftarrow \textcolor{red}{\text{ in standard form}}$