$x^2+y^2+ax+16=0$
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.
$x^2+y^2+ax+16=0$
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.
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
$$\frac{a^2}{4}-16 >0$$.
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 $a<-8$.