Let $\alpha,\beta$ be fixed complex numbers and $z$ is a variable complex number such that $|z-\alpha|^2+|z-\beta|^2=k.$Find out the condition for $k$ such that the locus of $z$ is a circle.
I think,if i take $\alpha,\beta$ as diametrically opposite points and $z$ on the circumference,then the $\alpha,\beta$ subtend right angle on $z$.But i cannot figure out the condition.The answer given is $k>\frac{1}{2}|\alpha-\beta|^2$
This answer is essentially taken from this duplicate question:
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Let $z=x+iy$ and $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$.
Then we have $|z-z_1|^2+|z-z_2|^2=k.$
This implies $$ |(x-x_1)+ i (y-y_1)|^2+|(x-x_2)+ i (y-y_2)|^2=k $$ $$ \therefore (x-x_1)^2+ (y-y_1)^2+(x-x_2)^2+ (y-y_2)^2=k $$ Expanding we get $$ 2x^2 +2y^2 -2x(x_1+x_2) -2y(y_1+y_2)=k-x_1^2 -y_1^2-x_2^2 -y_2^2 $$ $$\therefore x^2 +y^2 -x(x_1+x_2) -y(y_1+y_2)=\frac{1}{2}\left( k-x_1^2 -y_1^2-x_2^2 -y_2^2\right) $$ Now completing squares on the left hand side, $$ \Big(x-(\frac{x_1+x_1}{2})\Big)^2 +\Big(y-(\frac{y_1+y _1}{2})\Big)^2 = \frac{1}{2}\left(k-x_1^2 -y_1^2-x_2^2 -y_2^2\right) + \frac{1}{4}(x_1+x_2)^2+ \frac{1}{4}(y_1+y_2)^2$$
Expanding the right-hand side gives, $$\Big(x-(\frac{x_1+x_1}{2})\Big)^2 +\Big(y-(\frac{y_1+y _1}{2})\Big)^2 = \frac{1}{2}\left(k-x_1^2 -y_1^2-x_2^2 -y_2^2 + \frac{1}{2} x_1^2 + \frac{1}{2} x_2^2 +x_1x_2 + \frac{1}{2} x_1^2 + \frac{1}{2} y_2^2 +y_1y_2 \right) $$ and so $$\Big(x-(\frac{x_1+x_1}{2})\Big)^2 +\Big(y-(\frac{y_1+y _1}{2})\Big)^2 = \frac{1}{2}[k-\frac{1}{2}x_1^2 -\frac{1}{2}y_1^2-\frac{1}{2}x_2^2 -\frac{1}{2}y_2^2 +x_1x_2 +y_1y_2 ] $$ $$\Big(x-(\frac{x_1+x_1}{2})\Big)^2 +\Big(y-(\frac{y_1+y _1}{2})\Big)^2 = \frac{1}{2}[k-\frac{1}{2} [ (x_1 -x_2)^2 +(y_1-y_2)^2] ] $$ i.e. $$ \Big(x-(\frac{x_1+x_1}{2})\Big)^2 +\Big(y-(\frac{y_1+y _1}{2})\Big)^2 = \frac{1}{2}[k-\frac{1}{2} |z_1-z_2|^2 ] $$
This expression takes the same format at the equation of a circle if the right hand side $r^2$.
Thus, the term $k-\frac{1}{2} |z_1-z_2|^2$ should be strictly positive. That is, $k>\frac{1}{2} |z_1-z_2|^2$.