Algebra - Circles

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Section 3.3 : Circles

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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.