Given
$$ |z-a|^2 +|z-b|^2 = |a-b|^2 $$
where $(a,b,z=OP) $ are complex with a right angle at common point P in the complex plane. what is the curve it traces? Would the locus be also a circle on diameter $ |a-b|$?
Special case: when $(a,b) $ are real and of same sign then we have const $\tau$ blue circles and if of opposite sign then we have const $\sigma$ red circles as the required locus:
By the parallelogram law (or the equivalent median theorem) written for $\,z-a, z-b\,$:
$$ \begin{align} 2 \big|z-a\big|^2 + 2 \big|z-b\big|^2 &= \big|(z-a)-(z-b)\big|^2 + \big|(z - a) + (z - b)\big|^2 \\ &= \big|a-b\big|^2 + 4\, \bigg|z - \frac{a+b}{2}\bigg|^2 \end{align} $$
Under the given condition $\,|z-a|^2 + |z-b|^2 = |a-b|^2\,$, so:
$$ 2 \big|a-b\big|^2 = \big|a-b\big|^2 + 4\, \bigg|z - \frac{a+b}{2}\bigg|^2 \quad\implies \quad \bigg|z - \frac{a+b}{2}\bigg| = \bigg|\frac{a-b}{2}\bigg| $$
It follows that the locus is the circle having $\,a b\,$ as the diameter.
More generally, the same argument can be used to prove that the locus of $\,|z-a|^2+|z-b|^2 = c\,$ is a circle centered at the midpoint of $\,a b\,$.
Yet more generally, the locus of points with constant sum of squared distances to $\,n\,$ fixed points is a circle (in 2D, or a hypersphere in $\,\mathbb R^n$) centered at the centroid of those points.