Paul's Online Notes
Home / Algebra / Graphing and Functions / Circles
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### Section 3.3 : Circles

8. Determine the radius and center of the following circle. If the equation is not the equation of a circle clearly explain why not.

${x^2} + {y^2} + 8x + 20 = 0$

Show All Steps Hide All Steps

Start Solution

To do this problem we need to complete the square on the $$x$$ and $$y$$ terms. To help with this we’ll first get the number on the right side and group the $$x$$ and $$y$$ terms as follows.

${x^2} + 8x + {y^2} = - 20$ Show Step 2

Here is the number we need to complete the square for both $$x$$. Note that we don’t need to complete the square for the $$y$$.

$x:\,{\left( {\frac{8}{2}} \right)^2} = {\left( 4 \right)^2} = 16$ Show Step 3

Now, complete the square.

\begin{align*}{x^2} + 8x + 16 + {y^2} & = - 20 + 16\\ {\left( {x + 4} \right)^2} + {y^2} & = - 4\end{align*}

Don’t forget to add the number from Step 2 to both sides of the equation!

Show Step 4

Okay, at this point we can see that this equation is not the equation of a circle. The standard form of the circle is,

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

The right side is $${r^2}$$ and that must be a positive number (and the coefficients of the $$x$$ and $$y$$ must also be positive) which for our equation it is not. Therefore, this is not the equation of a circle.