**Complex locus of a circle**
**Yue Kwok Choy**
**(1) **It is easy to show that |z – z_{1}| = a , where z_{1}**C**, a**R** form a circle with centre P_{1}(z_{1}) and radius a , using an Argand Diagram.
**(2) **By putting z = x + yi and z_{1} = x_{1} + y_{1}i , we can transform the equation to well known Cartesian form : (x – x_{1})^{2} + (y – y_{1})^{2} = a^{2} . The equation, in fact, is a circle with centre (x_{1}, y_{1}) and radius a in the rectangular plane.
**(3) **Squaring the equation of circle in **(1)**, we get
|z – z_{1}|^{2} = a^{2}
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, z_{1} = x_{1} + y_{1}i in **(3)** gives back the Cartesian form of the circle.
**(5) **Putting z = r(cos + i sin ) , z_{1} = (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 z_{1} and z_{2} such that
P_{1}PP_{2} = is subtended by the chord at points on the arc,
using the s in the same segment theorem.
**(7) **Putting z = x + yi , z_{1} = x_{1} + y_{1}i , z_{2} = x_{2} + y_{2}i in **(6)**, we have :
**(8) **The last equation in **(7)** is a homogenous equation of degree 2 , also
**(a)** coeff. of x^{2} –term = coeff. of y^{2} 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 P_{1}P_{2} with the line segment P_{1}P_{2} removed.
**(b)** when = , the line segment P_{1}P_{2} .
**(c)** when 0 < < , an arc of a circle, terminating at P_{1} and P_{2} (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 P_{1} and P_{2} removed.
You may investigate the following loci :
**(a) ** or – , 0 < < .
**(b) ** , 0 < < .
**(11) **where z_{1} , z_{2} **C**, k > 0 , k 1 gives a circle (excluding points P_{1}(z_{1}), P_{2}(z_{2}) )
Note : When k = 1, the locus is the perpendicular bisector of the line joining P_{1}(z_{1}) and P_{2}(z_{2}) .
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 P_{1}(z_{1}) and P_{2}(z_{2}) ,
we have P_{1}P = k P_{2}P . This then reduces to a well-known geometry problem :
**The Circle of Apollonius**: Given two fixed points P_{1} and P_{2}, the locus of point P such that the ratio of P_{1}P to P_{2}P 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 P_{1}P_{2} internally and externally :
P_{1} A : AP_{2} = k : 1
P_{1} B : BP_{2} = –k : 1
By section formula:
A represents
B represents
The centre of the circle represents
and the radius = .
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