Q.Show that $\left|\dfrac{z-2}{z-3}\right|=2$ represents a circle. Find its centre and radius.

The key idea is to transform the given complex number equation into the standard Cartesian form of a circle equation, $x^2+y^2+2gx+2fy+c=0$.

Starting with the given equation:

$\left|\dfrac{z-2}{z-3}\right|=2$

This can be written as $|z-2| = 2|z-3|$.

Squaring both sides to eliminate the modulus:

$|z-2|^2 = 4|z-3|^2$

Substitute $z = x+iy$:

$|(x-2)+iy|^2 = 4|(x-3)+iy|^2$

$(x-2)^2 + y^2 = 4((x-3)^2 + y^2)$

$x^2 - 4x + 4 + y^2 = 4(x^2 - 6x + 9 + y^2)$

$x^2 - 4x + 4 + y^2 = 4x^2 - 24x + 36 + 4y^2$

Rearrange the terms to form the standard equation of a circle:

$0 = (4x^2 - x^2) + (4y^2 - y^2) + (-24x + 4x) + (36 - 4)$

$3x^2 + 3y^2 - 20x + 32 = 0$

Divide by 3:

$x^2 + y^2 - \dfrac{20}{3}x + \dfrac{32}{3} = 0$

Comparing this with the general circle equation $x^2+y^2+2gx+2fy+c=0$:

$2g = -\dfrac{20}{3} \implies g = -\dfrac{10}{3}$

$2f = 0 \implies f = 0$

$c_{const} = \dfrac{32}{3}$ …

The problem asks us to demonstrate that a given equation involving complex numbers represents a circle, and then to find its centre and radius. The core idea here is to understand what the modulus of a complex number signifies geometrically and then translate the given complex equation into its Cartesian equivalent.

The expression $|z-a|$ represents the distance between the complex number $z$ and the complex number $a$ in the Argand plane. So, the equation $\left|\dfrac{z-2}{z-3}\right|=2$ can be interpreted as a ratio of distances. Specifically, it means that the distance from $z$ to $2$ is twice the distance from $z$ to $3$.

When the ratio of distances from a point $z$ to two fixed points $A$ and $B$ is a constant $k \neq 1$, the locus of $z$ is a circle. This is a classic result known as Apollonius's Circle. Our task is to algebraically derive the standard Cartesian equation of a circle from this complex number relation.

Let's proceed step-by-step:

1. Isolate the moduli and interpret geometrically.

   The given equation is $\left|\dfrac{z-2}{z-3}\right|=2$.

   Using the property $|z_1/z_2| = |z_1|/|z_2|$, we can write this as:

   $\dfrac{|z-2|}{|z-3|} = 2$

   This implies:

   $|z-2| = 2|z-3|$

   Geometrically, this means that the distance from $z$ to the point $2+0i$ (which is $(2,0)$ in the Cartesian plane) is twice the distance from $z$ to the point $3+0i$ (which is $(3,0)$).

2. Convert to Cartesian coordinates.

   Let $z = x + iy$, where $x$ and $y$ are real numbers.

   Substitute $z = x+iy$ into the equation:

   $|(x+iy)-2| = 2|(x+iy)-3|$

   $|(x-2)+iy| = 2|(x-3)+iy|$

   Recall that for a complex number $a+bi$, its modulus is $|a+bi| = \sqrt{a^2+b^2}$. Applying this:

   $\sqrt{(x-2)^2 + y^2} = 2\sqrt{(x-3)^2 + y^2}$

3. Eliminate the square roots by squaring both sides.

   Squaring both sides simplifies the equation significantly:

   $\left(\sqrt{(x-2)^2 + y^2}\right)^2 = \left(2\sqrt{(x-3)^2 + y^2}\right)^2$

   $(x-2)^2 + y^2 = 4\left((x-3)^2 + y^2\right)$

   Watch out

   A common mistake is to forget to square the '2' on the right side, leading to $2((x-3)^2 + y^2)$ instead of $4((x-3)^2 + y^2)$. Always square the entire right-hand side.

4. Expand and simplify the equation.

   Expand the squared terms:

   $x^2 - 4x + 4 + y^2 = 4(x^2 - 6x + 9 + y^2)$

   Distribute the $4$ on the right side:

   $x^2 - 4x + 4 + y^2 = 4x^2 - 24x + 36 + 4y^2$

5. Rearrange into the general form of a circle equation.

   Move all terms to one side to set the equation to zero: …