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If $z_1$ and $z_2$ are fixed and satisfies $|z-z_1|^2+|z-z_2|^2=k$ what are the possible values of k so that this equation represents a circle?

I tried using pythagoras theorem,that the equation of circle should be $z-z_1|^2+|z-z_2|^2=|z_1-z_2|^2$.

So k should be $|z_1-z_2|^2$.But the answer is given is $k>\frac{1}{2}|z_1-z_2|^2$.Where am I going wrong?How to approach the problem (preferably geometrically) ?

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  • $\begingroup$ "I tried using pythagoras theorem,that the equation of circle should be $z-z_1|^2+|z-z_2|^2=|z_1-z_2|^2$." This is impossible to decipher, please explain what you did. $\endgroup$ – Did Oct 14 '15 at 15:17
  • $\begingroup$ How did you apply Pythagoras theroem ? $\endgroup$ – Nizar Oct 14 '15 at 15:19
  • $\begingroup$ Ok see let z1 and z2 be 2 points.Angle in a semicircle is 90 degrees.For z to lie on a circle whose diameter's end points are z1 and z2 then |z-z1|,|z-z2| and |z1-z2| are sides of a right triangle. $\endgroup$ – user220382 Oct 14 '15 at 15:21
  • $\begingroup$ who said that the diameter should have z1 and z2 as end points ? $\endgroup$ – Nizar Oct 14 '15 at 15:22
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    $\begingroup$ Ok so that was the mistake @Nizar.. $\endgroup$ – user220382 Oct 14 '15 at 15:24
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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 $|z-z_1|^2+|z-z_2|^2=k$ implies $$ |(x-x_1)+ i (y-y_1)|^2+|(x-x_2)+ i (y-y_2)|^2=k $$ $$ (x-x_1)^2+ (y-y_1)^2+(x-x_2)^2+ (y-y_2)^2=k $$ Expanding we get $$ 2x^2 +2y^2 -2x(x_1+x_2) -2y(y_1+y_2)=k-x_1^2 -y_1^2-x_2^2 -y_2^2 $$ divide by 2 we get $$x^2 +y^2 -x(x_1+x_2) -y(y_1+y_2)=\frac{1}{2}[k-x_1^2 -y_1^2-x_2^2 -y_2^2] $$ Complete squares on the left hand side $$ \Big(x-(\frac{x_1+x_1}{2})\Big)^2 +\Big(y-(\frac{y_1+y _1}{2})\Big)^2 = \frac{1}{2}[k-x_1^2 -y_1^2-x_2^2 -y_2^2] + \frac{1}{4}(x_1+x_2)^2+ \frac{1}{4}(y_1+y_2)^2$$ This implies $$\Big(x-(\frac{x_1+x_1}{2})\Big)^2 +\Big(y-(\frac{y_1+y _1}{2})\Big)^2 = \frac{1}{2}[k-x_1^2 -y_1^2-x_2^2 -y_2^2 + \frac{1}{2} x_1^2 + \frac{1}{2} x_2^2 +x_1x_2 + \frac{1}{2} x_1^2 + \frac{1}{2} y_2^2 +y_1y_2 ] $$ 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. $$ \Big(x-(\frac{x_1+x_1}{2})\Big)^2 +\Big(y-(\frac{y_1+y _1}{2})\Big)^2 = \frac{1}{2}[k-\frac{1}{2} |z_1-z_2|^2 ] $$ Note that this has the well known equation of a circle just if having on the right hand side $r^2$, and so the term $k-\frac{1}{2} |z_1-z_2|^2$ should be positive strictly i.e. $k>\frac{1}{2} |z_1-z_2|^2$.

Small Remark : If you notice, always the center of your circle is the midpoint between $z_1$ and $z_2$. Special case when $k=|z_1-z_2|^2$, the two points $z_1$ and $z_2$ are on the circle .

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    $\begingroup$ Well done :-).. $\endgroup$ – user220382 Oct 14 '15 at 15:53
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$$|z-z_1|^2+|z-z_2|^2=k\tag{1}$$ let $z=w+\dfrac{z_1+z_2}{2}$ then $(1)$ becomes $$|w+\dfrac{z_1-z_2}{2}|^2+|w-\dfrac{z_1-z_2}{2}|^2=k$$ or $$|w+\alpha|^2+|w-\alpha|^2=k\tag{2}$$ where $\alpha=\dfrac{z_1-z_2}{2}$, from $(2)$ $$2w\overline{w}+2\alpha\overline{\alpha}=k$$ or $$|w|^2=\dfrac{k}{2}-|\alpha|^2$$ this is the equation of circle $|w|=r$ with $r=\sqrt{\dfrac{k}{2}-|\alpha|^2}$ if $\dfrac{k}{2}-|\alpha|^2>0$ or $$\color{blue}{k>2|\alpha|^2=\dfrac12|z_1-z_2|^2}$$

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