In the equation $x^2+\mathrm{px}+q=0$ if we fix $q$ and vary $p$ and observe the complex solutions which occur (for $p^2-4q<0$) then, as $p$ changes, the complex roots of the equation also change. If these roots are plotted on an Argand diagram then you will see that the roots lie on a circle.

To prove that the complex roots do lie on a circle we use the fact that the product of the roots of the quadratic equation $x^2+\mathrm{px}+q=0$ is $q$.

This equation has roots $z_1$ and $z_2$ given by

${\displaystyle \frac{-p\pm \sqrt{p^2-4q}}{2}}$

so

${\displaystyle z_1=\frac{-p+\sqrt{p^2-4q}}{2}=u+\mathrm{iv}}$

and

${\displaystyle z_2=\frac{-p-\sqrt{p^2-4q}}{2}=u-\mathrm{iv}}$

where $u+\mathrm{iv}$ and $u-\mathrm{iv}$ are complex conjugates.

The product of the complex conjugate roots is given by

${\displaystyle z_1{z}_2=(u+\mathrm{iv})(u-\mathrm{iv})=u^2+v^2=q.}$

Hence as the quadratic changes keeping $q$ fixed and varying $p$ the locus of the complex roots in the $(u,v)$ plane is the circle with radius $\sqrt{q}$ centre at the origin.