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Suppose that $k|z-z_1|=l|z-z_2|$ where $k\neq l$ and both are positive real numbers.

Show that the locus of $z$ in the Argand diagram is a circle with center: $$\frac{k^2 z_1-l^2 z_2}{k^2-l^2}$$ and radius: $$\frac{kl|z_2-z_1|}{|k^2-l^2|}$$ by the Geometric Means method.

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closed as off-topic by Nosrati, Wouter, Cesareo, Brahadeesh, user91500 Sep 17 '18 at 9:31

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  • $\begingroup$ What is the Geometric Means method? $\endgroup$ – Gerry Myerson Mar 5 '13 at 2:44
  • $\begingroup$ Maybe it is by geometric means. $\endgroup$ – André Nicolas Mar 5 '13 at 2:53
  • $\begingroup$ Yes by geometric means but not by Cartesian form of z. $\endgroup$ – user65046 Mar 5 '13 at 3:30
  • $\begingroup$ @user65046: You can merge your current account with the one you used to post this question originally by going here: From any page footer -> 'contact us' >> 'Merge user profiles' $\endgroup$ – Zev Chonoles Mar 5 '13 at 3:41
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You can use $|z|^2=z\bar{z}$ to solve this problem. In fact, from $k|z−z_1|=l|z−z_2|$, we have $k^2|z−z_1|^2=l^2|z−z_2|^2$ which implies $k^2(z−z_1)(\bar{z}−\bar{z}_1)=l^2(z−z_2)(\bar{z}-\bar{z}_2)$ or $$ k^2(|z|^2-z\bar{z}_1-\bar{z}z_1+|z_1|^2)=l^2(|z|^2-z\bar{z}_2-\bar{z}z_2+|z_2|^2). $$ Thus we obtain $$ (k^2-l^2)|z|^2-z(k^2\bar{z}_1-l^2\bar{z}_2)-\bar{z}(k^2z_1-l^2z_2)+k^2|z_1|^2-l^2|z_2|^2=0 $$ from which we derive $$ |z|^2-z\frac{k^2\bar{z}_1-l^2\bar{z}_2}{k^2-l^2}-\bar{z}\frac{k^2z_1-l^2z_2}{k^2-l^2}=-\frac{k^2|z_1|^2-l^2|z_2|^2}{k^2-l^2}. $$ So $$ |z-\frac{k^2z_1-l^2z_2}{k^2-l^2}|^2=\left|\frac{k^2z_1-l^2z_2}{k^2-l^2}\right|^2-\frac{k^2|z_1|^2-l^2|z_2|^2}{k^2-l^2} $$ whose the RHS is $(\frac{kl|z_1-z_2|}{|k^2-l^2|})^2$ (check!!). So $$ |z-\frac{k^2z_1-l^2z_2}{k^2-l^2}|=\frac{kl|z_1-z_2|}{|k^2-l^2|}. $$ It says that this is a circle with center $\frac{k^2z_1-l^2z_2}{k^2-l^2}$ and radius $\frac{kl|z_1-z_2|}{|k^2-l^2|}$.

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