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The problem is : if $z$ lies on a circle with diameter having endpoints $z_1$ and $z_2$ then show that $|z-z_1|^2 + |z-z_2|^2 = |z_1 - z_2|^2$ where $z, z_1, z_2 \in \mathbb{C}$.

The angle subtended by the diameter on any point on the circle is a right angle and thus $|z-z_1|$, $|z-z_2|$ and $|z_1-z_2|$ are the lengths of a right-angled triangle. The equation above then follows from the Pythagorean Theorem.

Now the equation for $z$ can also be written as $\left|z - \left(\dfrac{z_1+z_2}{2}\right)\right| = \dfrac{|z_1-z_2|}{2}$ since $\left(\dfrac{z_1+z_2}{2}\right)$ is the centre and $\dfrac{|z_1-z_2|}{2}$ is the radius of the circle.

But since the locus of both the equations are the same, I figured that it should be possible to prove them equal. So here's what I did:

$$\begin{align} \left|z - \left(\dfrac{z_1+z_2}{2}\right)\right| = \dfrac{|z_1-z_2|}{2} &\iff |2z - (z_1+z_2)| =|z_1-z_2| \\ &\iff |(z - z_1)+ (z-z_2)|^2=|z_1-z_2|^2 \end{align}$$

Using $|z_1+z_2|^2=|z_1|^2 + |z_2|^2 + \Re{(z_1\overline{z_2})}$,

$|z - z_1|^2+|z-z_2|^2 + \Re{((z-z_1)\overline{(z-z_2)})}=|z_1-z_2|^2$. Comparing it with what I have to show, it seems like I have to prove $\Re{((z-z_1)\overline{(z-z_2)})} = 0$, but I can't think of a way of doing that.

Thank you in advance!

Edit: To sum it up, my question is how to show that $$\left|z - \left(\dfrac{z_1+z_2}{2}\right)\right| = \dfrac{|z_1-z_2|}{2}\iff|z-z_1|^2 + |z-z_2|^2 = |z_1 - z_2|^2$$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,$where $z, z_1, z_2 \in \mathbb{C}$.

Alraxite
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3 Answers3

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If $z=x+iy,z_k=x_k+iy_k$ for $k=1,2$

the real part of $(z-z_1)(\overline{z-z_2})$ is $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)$

Using Article $145$ The elements of coordinate geometry by Loney

$$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$$

lab bhattacharjee
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After reading the comment, the question seems to be

If $z_1$ and $z_2$ are at different ends of the diameter of a circle, and $z$ is also on that circle, prove $$ \mathrm{Re}((z-z_1)(\overline{z-z_2}))=0 $$

Method 1: One way to see this is to note that $\mathrm{Re}(z\bar{w})=z\cdot w$ (dot product as vectors). Since $z-z_1$ and $z-z_2$ are perpendicular, $(z-z_1)\cdot(z-z_2)=0$.

Method 2: Another way of looking at this, is to let $c$ be the center of the circle and let $w=z-c$, $w_1=z_1-c$, and $w_2=z_2-c$. Then $w_2=-w_1$ and $|w|=|w_1|=|w_2|=r$ $$ \begin{align} (z-z_1)(\overline{z-z_2}) &=(w-w_1)(\overline{w-w_2})\\ &=w\bar{w}+w_1\bar{w}_2-w\bar{w}_2-\bar{w}w_1\\ &=w\bar{w}-w_1\bar{w}_1+(w\bar{w}_1-\bar{w}w_1)\\ &=r^2-r^2+2i\,\mathrm{Im}(w\bar{w}_1)\\ &=2i\,\mathrm{Im}(w\bar{w}_1) \end{align} $$ which is pure imaginary. Therefore, $$ \mathrm{Re}((z-z_1)(\overline{z-z_2}))=0 $$

robjohn
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  • I know that this can be shown using this theorem. I'm trying to prove that $|z - (\dfrac{z_1+z_2}{2})| = \dfrac{|z_1-z_2|}{2}$ is the same as $|z-z_1|^2 + |z-z_2|^2 = |z_1 - z_2|^2$. I think my question is unclear. – Alraxite Jun 30 '13 at 15:55
  • I think your question's clear, @Alraxite : what you want in your comment and also in the comment below my answer is what is unclear since you did not ask this equivalence...which, as already pointed out below my answer, is trivial. – DonAntonio Jun 30 '13 at 16:21
  • @Alraxite: I hope this more closely addresses your question. – robjohn Jun 30 '13 at 16:41
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WLOG suppose we have the circle $|z|=r\;$ and that $\,z_1=-r\;,\;\;z_2=r\;,\;z=re^{it}\;,\;\;0<t<\pi\;$ , then

$$|z-z_1|^2+|z-z_2|^2=|r(\cos t+1)+ir\sin t|^2+|r(\cos t-1)+i\sin t|^2=$$

$$=r^2(\cos^2t+2\cos t+1)+r^2\sin^2t+r^2(\cos^2t-2\cos t+1)+r^2\sin^2t=$$

$$=4r^2=|z_1-z_2|^2$$

DonAntonio
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  • But how does this show $|z - (\dfrac{z_1+z_2}{2})| = \dfrac{|z_1-z_2|}{2} \iff |z-z_1|^2 + |z-z_2|^2 = |z_1 - z_2|^2$? – Alraxite Jun 30 '13 at 15:53
  • I've no idea what you're asking in your comment, @Alraxite : I addressed and answered you post's title, which is the right side of the expression you wrote in your comment above...but after taking a peek at it I think your left side is trivial: it says the distance from $,z,$ to the circle's center equals the circle's radius...which, again, is trivial since $;z;$ belongs to the circle! – DonAntonio Jun 30 '13 at 16:17
  • Let me explain. The problem is 'if $z$ lies on a circle with diameter having endpoints $z_1$ and $z_2$ then show that $|z-z_1|^2 + |z-z_2|^2 = |z_1 - z_2|^2$ where $z, z_1, z_2 \in \mathbb{C}$'. The part '$z$ lies on a circle with diameter having endpoints $z_1$ and $z_2$' is the same as saying $\left|z - \left(\dfrac{z_1+z_2}{2}\right)\right| = \dfrac{|z_1-z_2|}{2}$ for obvious reasons. So the new problem I was trying to solve was the one that I stated in my first comment. – Alraxite Jun 30 '13 at 16:43
  • Please note that I use no other assumptions. That is, I'm trying to prove $ |z-z_1|^2 + |z-z_2|^2 = |z_1 - z_2|^2$ using only the information provided by $|z - (\dfrac{z_1+z_2}{2})| = \dfrac{|z_1-z_2|}{2}$. The point is: I've changed the original assumption given in the problem to this. Anyway, I do appreciate your answer since it shows me a different way to solve the original problem. – Alraxite Jun 30 '13 at 16:44