# How do you find the standard form given x^2+y^2-8x+4y+11=0?

Aug 2, 2017

${\left(x - 4\right)}^{2} + {\left(y + 2\right)}^{2} = 9$

#### Explanation:

Firstly, which conic is this?

Well, the coefficients in front of the ${x}^{2}$ and ${y}^{2}$ terms are the same, and there is a plus sign between them, so it is a circle.

The standard form equation of a circle is given by

ul((x-h)^2 + (y-k)^2 = r^2

where

• $h$ is the $x$-coordinate of the center of the circle

• $k$ is the $y$-coordinate of the center of the circle

• $r$ is the radius of the circle

We'll need to complete the square to find the standard form equation.

Let's move the constant $11$ to the right hand side, and group $x$- and $y$-terms together:

${x}^{2} - 8 x + {y}^{2} + 4 y = - 11$

Set up the completion of the square for $x$ and $y$:

$\left({x}^{2} - 8 x + \underline{\text{ ")) + (y^2 + 4y + ul(" }}\right) = - 11$

We're looking for the "$c$" term in the parabolic equation, and to find that we do

ul(c = (b/2)^2

or

ul(c = (b^2)/4

The $b$ term for the $x$ is $- 8$ (the variable in front of the linear term $x$), so we have

x{color(white)(a)c = ((-8)^2)/4 = color(red)(ul(16

Similarly, for the $y$, $b = 4$, so we have

y{color(white)(a)c = (4^2)/4 = color(green)(ul(4

Now, we plug these values into the equation:

$\left({x}^{2} - 8 x + \textcolor{red}{16}\right) + \left({y}^{2} + 4 y + \textcolor{g r e e n}{4}\right) = - 11$

Now, we add both values of $c$ to the constant on the right side:

$\left({x}^{2} - 8 x + \textcolor{red}{16}\right) + \left({y}^{2} + 4 y + \textcolor{g r e e n}{4}\right) = - 11 + \textcolor{red}{16} + \textcolor{g r e e n}{4}$

Each of the terms can now be represented as the square of a monomial, and so we have

color(blue)(ulbar(|stackrel(" ")(" "(x-4)^2 + (y+2)^2 = 9" ")|)

(to get the value of $h$ and $k$, simply take half of the $b$ value for each variable; i.e. ${\overbrace{- \frac{8}{2} = - 4}}^{x}$ and ${\overbrace{\frac{4}{2} = 2}}^{y}$

For this equation here, we can determine:

• the center point $\left(h , k\right)$ of the circle is $\left(4 , - 2\right)$

• the radius $r$ is $\sqrt{9} = 3$