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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.

\[{x^2} + {y^2} + 8x + 20 = 0\]

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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.