# What is the standard form of the equation of a circle with center at (3, 2) and through the point (5, 4)?

Mar 11, 2016

${\left(x - 3\right)}^{2} + {\left(y - 2\right)}^{2} = 8$

#### Explanation:

The standard form of the equation of a circle is :

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

where (a , b ) are the coords of centre and r , the radius.

Here the centre is known but require to find radius. This can be done using the 2 coord points given.

using the$\textcolor{b l u e}{\text{ distance formula }}$

$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

let$\left({x}_{1} , {y}_{1}\right) = \left(3 , 2\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(5 , 4\right)$

$d = r = \sqrt{{\left(5 - 3\right)}^{2} + {\left(4 - 2\right)}^{2}} = \sqrt{8}$

equation of circle is $: {\left(x - 3\right)}^{2} + {\left(y - 2\right)}^{2} = {\left(\sqrt{8}\right)}^{2}$