I have been informed that on March 7th from 6:00am to 6:00pm Central Time Lamar University will be doing some maintenance to replace a faulty UPS component and to do this they will be completely powering down their data center.
Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
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 SolutionTo 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 2Here 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 3Now, 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 4Okay, 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.