Complex locus of a circle

Complex locus of a circle

Yue Kwok Choy
(1) It is easy to show that |z – z₁| = a , where z₁C, aR form a circle with centre P₁(z₁) and radius a , using an Argand Diagram.
(2) By putting z = x + yi and z₁ = x₁ + y₁i , we can transform the equation to well known Cartesian form : (x – x₁)² + (y – y₁)² = a² . The equation, in fact, is a circle with centre (x₁, y₁) and radius a in the rectangular plane.
(3) Squaring the equation of circle in (1), we get

|z – z₁|² = a² 

We get another form of circle: , aC, cR .

Here in order not to get an imaginary or degenerate circle .

(4) Putting z = x + yi, z₁ = x₁ + y₁i in (3) gives back the Cartesian form of the circle.
(5) Putting z = r(cos  + i sin ) , z₁ = (cos  + i sin ) , ( ,  are constants) in (3) :

we get the polar form of a circle :

, with centre = (, ) and radius = .




(6) , 0 <  <  gives an arc and not a circle.

As in the figure, the locus gives an arc of the circle standing

on the chord with end points z₁ and z₂ such that

P₁PP₂ =  is subtended by the chord at points on the arc,

using the s in the same segment theorem.
(7) Putting z = x + yi , z₁ = x₁ + y₁i , z₂ = x₂ + y₂i in (6), we have :







(8) The last equation in (7) is a homogenous equation of degree 2 , also

(a) coeff. of x² –term = coeff. of y² term and

(b) there is no xy-term,

It therefore gives a complete circle and not an arc.

The "problematic" step in (7) marked by "" changes the arc into a circle .
(9) The locus of P in (7) represents :

(a) when  = 0 , the whole line P₁P₂ with the line segment P₁P₂ removed.

(b) when  =  , the line segment P₁P₂ .

(c) when 0 <  < , an arc of a circle, terminating at P₁ and P₂ (and excluding these points)

(d) when  = /2 , a semicircle and the supplementary semicircle is given by  = 3/2 .
(10) or  +  , 0 <  <  gives a complete circle with P₁ and P₂ removed.

You may investigate the following loci :

(a) or  –  , 0 <  <  .

(b) , 0 <  <  .
(11) where z₁ , z₂ C, k > 0 , k  1 gives a circle (excluding points P₁(z₁), P₂(z₂) )

Note : When k = 1, the locus is the perpendicular bisector of the line joining P₁(z₁) and P₂(z₂) .

Proof :

Comparing this with that given in (3), we get a circle with centre and radius a, where

on simplification (exercise)
(12) and if we take P(z) a variable point and P₁(z₁) and P₂(z₂) ,

we have P₁P = k P₂P . This then reduces to a well-known geometry problem :

The Circle of Apollonius: Given two fixed points P₁ and P₂, the locus of point P such that the ratio of P₁P to P₂P is constant , k, is a circle.
The Circle of Apollonius is not discussed here. Interested readers may consult web-sites such as:

http://jwilson.coe.uga.edu/emt725/Apollonius/Cir.html
If we know that the locus is a circle, then finding the centre and radius is easier.

As in the diagram, C is the centre and AB is the diameter of the circle.

Then A and B divide P₁P₂ internally and externally :

P₁ A : AP₂ = k : 1

P₁ B : BP₂ = –k : 1

 By section formula:

A represents

B represents

 The centre of the circle represents

and the radius = .