# How do you find the equation of the circle given Center at the origin; r=squareroot 5?

Sep 1, 2016

${x}^{2} + {y}^{2} = 5$

#### Explanation:

The standard form of the equation of a circle is.

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

here the centre is at the origin, hence a = 0 , b = 0
and $r = \sqrt{5}$

substitute into the standard equation.

${\left(x - 0\right)}^{2} + {\left(y - 0\right)}^{2} = {\left(\sqrt{5}\right)}^{2} \text{ simplifying to}$

${x}^{2} + {y}^{2} = 5 \text{ the equation of the circle}$

However, it is worth noting and remembering that when the circle is centred at the origin, the standard equation reduces to.
$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{x}^{2} + {y}^{2} = {r}^{2}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$