Question

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

Answer 1

Hint: At first take cross multiplication and then take z as $x+iy$ and then use form that if $w=a+ib$ then $\left| w \right|={{a}^{2}}+{{b}^{2}}$. After solving and finding equation of circle convert it in form ${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}=c^2$ where (a, b) is centre and c is radius.

Complete step-by-step answer:
In the question we are given an equation $\left| \dfrac{z-2}{z-3} \right|=2$ represents a circle and we have to find it’s centre and radius.
Here z is a complex number.
Before doing so, we will learn what complex numbers are.
A complex number is a number that can be written in form of $a+bi$ where a, b are real numbers and I is a solution of the equation ${{x}^{2}}=-1.$ This is because no real value satisfies for equation ${{x}^{2}}+1=0$ or ${{x}^{2}}=-1$, hence I is called imaginary numbers. For the complex number $a+ib$, a is considered as the real part and b as imaginary part. Despite the historical nomenclature “imaginary” , complex numbers are regarded in the mathematical sciences as just as “real” as real numbers, and are fundamental in any aspect of scientific description of the natural word.
So, given equation is
$\left| \dfrac{z-2}{z-3} \right|=2$
On cross multiplication we get,
$\left| z-2 \right|=2\left| z-3 \right|$
As z is a complex number then z can be taken as a $x+iy$ where x, y are real numbers.
So on substituting $z=x+iy$ we get,
$\left| x+iy-2 \right|=2\left| x+iy-3 \right|$
The equation can be rearranged and written as,
$\left| (x-2)+iy \right|=2\left| (x-3)+iy \right|$
Now we will apply formula that
If $w=a+ib$ then $\left| w \right|=\sqrt{{{a}^{2}}+{{b}^{2}}}$
So,
$\left| x-2+iy \right|=\sqrt{{{(x-2)}^{2}}+{{y}^{2}}}$ and $\left| x-3+iy \right|=\sqrt{{{(x-3)}^{2}}+{{y}^{2}}}$
So we can rewrite equation as,
$\sqrt{{{(x-2)}^{2}}+{{y}^{2}}}=2\sqrt{{{(x-3)}^{2}}+{{y}^{2}}}$
Now we will square on both sides we will get,
{{(x-2)}^{2}}+{{y}^{2}}=4\left\\{ {{(x-3)}^{2}}+{{y}^{2}} \right\\}
Now we will break and do the expansions we get,
{{x}^{2}}-4x+4+{{y}^{2}}=4\left\\{ \left( {{x}^{2}}-6x+9 \right)+{{y}^{2}} \right\\}
On simplifying the equation we get,
${{x}^{2}}-4x+4+{{y}^{2}}=4{{x}^{2}}-24x+36+4{{y}^{2}}$
Now taking all variables and constants in one side of equation we will get,
$3{{x}^{2}}-20x+32+3{{y}^{2}}=0$
Now we will divide the whole equation of 3 we will get
${{x}^{2}}-\dfrac{20}{3}x+\dfrac{32}{3}+{{y}^{2}}=0$
Now we will write the equation as,
${{x}^{2}}-2\left( \dfrac{10}{3} \right)x+{{\left( \dfrac{10}{3} \right)}^{2}}-\dfrac{4}{9}+{{y}^{2}}=0$
So we can write the equation as,
${{\left( x-\dfrac{10}{3} \right)}^{2}}+{{\left( y-0 \right)}^{2}}=\dfrac{4}{9}={{\left( \dfrac{2}{3} \right)}^{2}}$
Now as we know that if equation of circle is ${{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{1}} \right)}^{2}}={{r}^{2}}$ then its centre is $\left( {{x}_{1}},{{y}_{1}} \right)$ and radius is ‘r’ so if the equation is
${{\left( x-\dfrac{10}{3} \right)}^{2}}+{{\left( y-0 \right)}^{2}}={{\left( \dfrac{2}{3} \right)}^{2}}$
Then the centre will be $\left( \dfrac{10}{3},0 \right)$ and radius will be $\dfrac{2}{3}$ .
Hence it’s centre is $\left( \dfrac{10}{3},0 \right)$ and radius is $\left( \dfrac{2}{3} \right)$ .

Note: Also we can find centre and radius of circle, given equation if it is in form ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ then it’s centre will be (-g, -f) and radius will be $\sqrt{{{g}^{2}}+{{f}^{2}}-c}$ .