# How do you find the radius of a circle with the equation x^2 + y^2 + 4x - 8y + 13 = 0?

May 3, 2018

$\text{radius } = \sqrt{7}$

#### Explanation:

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

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

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

$\text{to obtain this form "color(blue)"complete the square"" on both}$
$\text{the x and y terms}$

${x}^{2} + 4 x + {y}^{2} - 8 y = - 13$

$\Rightarrow {x}^{2} + 2 \left(2\right) x \textcolor{red}{+ 4} + {y}^{2} + 2 \left(- 4\right) y \textcolor{m a \ge n t a}{+ 16} = - 13 \textcolor{red}{+ 4} \textcolor{m a \ge n t a}{+ 16}$

$\Rightarrow {\left(x + 2\right)}^{2} + {\left(y - 4\right)}^{2} = 7 \leftarrow \textcolor{b l u e}{\text{in standard form}}$

$\Rightarrow {r}^{2} = 7 \Rightarrow r = \sqrt{7}$

May 3, 2018

$r = \sqrt{7}$

#### Explanation:

Given: circle equation ${x}^{2} + {y}^{2} + 4 x - 8 y + 13 = 0$

The standard form of circle equation is: ${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$

where center$= \left(h , k\right)$, $\text{ } r =$radius

Use completing of the square to find the equation of the circle:

First combine the $x$ terms and the $y$ terms and put the constant(s) on the right side of the equation:

$\left({x}^{2} + 4 x\right) + \left({y}^{2} - 8 y\right) = - 13$

To complete the square take $\frac{1}{2}$ of the $x$-term $= \frac{1}{2} \cdot 4 = 2$ and $\frac{1}{2}$ of the $y$-term $= \frac{1}{2} \cdot - 8 = - 4$.

Add the extra added to the left side when you complete the square (${2}^{2}$ and (-4)^2) to the right side:

${\left(x + 2\right)}^{2} + {\left(y - 4\right)}^{2} = - 13 + {2}^{2} + {\left(- 4\right)}^{2}$

Simplify:

${\left(x + 2\right)}^{2} + {\left(y - 4\right)}^{2} = 7$

Since ${r}^{2} = 7 , \text{ } r = \sqrt{7}$