Question

Find the circumcentre of the triangle whose vertices are given by the complex numbers $z_1,z_2,z_3$.

• A. $z=\frac{\sum z_1\overline{z_1}(z_2-z_3)}{\sum \overline{z_1}(z_2-z_3)}$
• B. $z=\frac{z_2{\overline{z}}_2(z_1-z_3)}{\sum {\overline{z}}_2(z_1-z_2)}$
• C. $z=\frac{z_3{\overline{z}}_3(z_1-z_2)}{\sum {\overline{z}}_3(z_1-z_3)}$
• D. none of these

Answer 1

The correct option is A $z=\frac{\sum z_1\overline{z_1}(z_2-z_3)}{\sum \overline{z_1}(z_2-z_3)}$

Let C be the circum-center. Distance of this point from the vertices are equal.
Thus, $|z_1-c|=|z_2-c|=|z_3-c|$
Hence, $(z_1-c)(\overline{z_1}-\overline{c})=(z_2-c)(\overline{z_2}-\overline{c})=(z_3-c)(\overline{z_3}-\overline{c})$
From the first 2 terms, we get:
$-c\overline{z_1}-z_1\overline{c}+z_1\overline{z1}=-c\overline{z_2}-z_2\overline{c}+z_2\overline{z_2}$
=> $(z_1-z_2)\overline{c}+(\overline{z_1}-\overline{z_2})c=z_1\overline{z_1}-z_2\overline{z_2}$
=> $(z_1-z_2)\overline{c}+(\overline{z_1}-\overline{z_2})c=z_1\overline{z_1}-z_2\overline{z_2}$
Similarly, from the last 2 terms:
$(z_2-z_3)\overline{c}+(\overline{z2}-\overline{z_3})c=z_2\overline{z_2}-z_3\overline{z_3}$
Solving for C:
$c=\frac{\sum z_1\overline{z_1}(z_2-z_3)}{\sum \overline{z_1}(z_2-z_3)}$
Hence, (A) is correct.