Equation of a Circle:
Suppose a fixed complex number z0 and take z be any complex number which moves in such a path that it's distance from z0 is always same to 'r'. That implies z could lie on a circle whose centre is z0 and radius r. And it's equation could be |z -z0| = r .
, where centre = -a and radius = ![310_Equation of a Circle2.png](https://www.expertsmind.com/../CMSImages/310_Equation of a Circle2.png)
Example : If z1, z2, z3 are complex numbers such that
, denotes that the points shown by z1, z2, z3 lie on a circle passing through the origin.
Solution: Since P(z1), Q(z2), R(z3) and S(z4) are concyclic points,
![1244_Equation of a Circle3.png](https://www.expertsmind.com/../CMSImages/1244_Equation of a Circle3.png)
![931_Equation of a Circle4.png](https://www.expertsmind.com/CMSImages/931_Equation%20of%20a%20Circle4.png)
=>1/2 = real, which is true.
Hence z1, z2, z3 and the origin are concyclic.
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