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$x^2+y^2+ax+16=0$

Determine for which value of $a$ the equation represents a circle.

How would one tackle such a problem? I have no experience with these types of questions.

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3  
Complete the square such that $x^2$ and $ax$ both result from the expansion of $(x-\beta)^2$ for some $\beta$ that depends on $a$. –  Henning Makholm Oct 14 '12 at 15:16

2 Answers 2

up vote 3 down vote accepted

To become the equation of a circle, you need to get rid of $ax$, and this is a pretty standard completing the square problem:

$$x^2+ax+(\frac{a}{2})^2+y^2+16= (\frac{a}{2})^2$$

or

$$(x+\frac{a}{2})^2+y^2= \frac{a^2}{4}-16$$

This is the equation of a circle if and only if the RHS is positive. Solve now

$$\frac{a^2}{4}-16 >0$$.

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You light me. Thanks for noting my mistake (+1). –  Babak S. Oct 14 '12 at 15:34
    
@BabakSorouh no problem, we all do mistakes sometimes ;) –  N. S. Oct 14 '12 at 15:35
    
Thank you sir, I understand. –  JohnHedge Oct 14 '12 at 15:36

We complete the square to get: $$\left(x+\frac{a}{2}\right)^{2}+y^{2}-\frac{a^{2}}{4}+16=0$$ Thus $$\left(x+\frac{a}{2}\right)^{2}+y^{2}=\frac{a^2}{4}-16$$ Which is a circle with centre $(-a/2,0)$ and radius $\frac{a^{2}}{4}-16$. This only makes sense when $\frac{a^{2}}{4}-16>0$, so $a^{2}>64$, giving $a>8$ or $a<-8$.

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