Complex locus of a circle



Yüklə 17,65 Kb.
tarix17.11.2018
ölçüsü17,65 Kb.

Complex locus of a circle

Yue Kwok Choy
(1) It is easy to show that |z – z1| = a , where z1C, 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 , z2C, 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 = .






Dostları ilə paylaş:


Verilənlər bazası müəlliflik hüququ ilə müdafiə olunur ©genderi.org 2017
rəhbərliyinə müraciət

    Ana səhifə