Prove that the locus $|z-z_1|^2 +|z-z_2|^2=k$ of a moving point $z$ on the Argand plane is a circle when $|z_1-z_2|^2\leqslant 2k$.

What i have tried out- $|z-1|^2+|z-2|^2=k$ let $z=x+iy$ $$\\ \implies2x^2-6x+5+2y^2-k=0$$ This is the equation of a circle. centre $=(\frac32,0)$ radius=$\sqrt{\dfrac{(2k−1)}4 }$ means $\\ k\geqslant\frac12$


Try with $z=x+iy, z_1=a+ib, z_2=c+id$

On rearrangement we have, $$x^2+y^2-x(a+c)-y(b+d)+\frac{a^2+b^2+c^2+d^2-k}2=0$$

$$\implies \left(x-\frac{a+c}2\right)^2+\left(y-\frac{b+d}2\right)^2=\frac{2k-2a^2-2b^2-2c^2-2d^2+(a+c)^2+(b+d)^2}4=\frac{2k-\{(a-c)^2+(b-d)^2\}}4$$

For real circle we need $2k-\{(a-c)^2+(b-d)^2\}\ge0$

But here, $|z_1-z_2|=?$


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