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Question: If the equation \[\dfrac{\left| z-2 \right|}{\left| z-3 \right|}=2\] represents a circle, then the r...

If the equation ∣z−2∣∣z−3∣=2\dfrac{\left| z-2 \right|}{\left| z-3 \right|}=2 represents a circle, then the radius is equal to
1. 1
2. 13\dfrac{1}{3}
3. 34\dfrac{3}{4}
4. 23\dfrac{2}{3}

Explanation

Solution

This question involves the concept of complex numbers in this question, we have to put z=x+iyz=x+iy and solve using this concept likez=∣z∣=x2+y2z=\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}, if any complex number then also we have to write in this form only. So by substituting these in the expression which is given in question to find the radius r.

Complete step-by-step solution:
In this particular problem expression is given that
∣z−2∣∣z−3∣=2\dfrac{\left| z-2 \right|}{\left| z-3 \right|}=2
So the above equation can be simplified further we get:
∣z−2∣=2∣z−3∣\left| z-2 \right|=2\left| z-3 \right|
Apply the concept of a complex number that is z=x+iyz=x+iy and substitute this value in the above equation we get:
∣x+iy−2∣=2∣x+iy−3∣\left| x+iy-2 \right|=2\left| x+iy-3 \right|
After rearranging the term in the above equation we get:
∣(x−2)+iy∣=2∣(x−3)+iy∣\left| (x-2)+iy \right|=2\left| (x-3)+iy \right|
By applying the formula of a complex number that is z=∣z∣=x2+y2z=\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}} substitute the value in the above equation.
(x−2)2+y2=2(x−3)2+y2\sqrt{{{(x-2)}^{2}}+{{y}^{2}}}=2\sqrt{{{(x-3)}^{2}}+{{y}^{2}}}
Squaring this above equation on both sides we get:
(x−2)2+y2=4[(x−3)2+y2]{{(x-2)}^{2}}+{{y}^{2}}=4\left[ {{(x-3)}^{2}}+{{y}^{2}} \right]
Apply the property of (a−b)2=a2+b2−2ab{{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab and substitute in this above equation we get:
x2−4x+4+y2=4(x2−6x+9+y2){{x}^{2}}-4x+4+{{y}^{2}}=4\left( {{x}^{2}}-6x+9+{{y}^{2}} \right)
Multiply 4 on inside the bracket we get:
x2−4x+4+y2=4x2−24x+36+4y2{{x}^{2}}-4x+4+{{y}^{2}}=4{{x}^{2}}-24x+36+4{{y}^{2}}
By rearranging the term we get:
x2−4x+4+y2−4x2+24x−36−4y2=0{{x}^{2}}-4x+4+{{y}^{2}}-4{{x}^{2}}+24x-36-4{{y}^{2}}=0
By simplifying further and solving further we get:
−3x2+20x−32−3y2=0-3{{x}^{2}}+20x-32-3{{y}^{2}}=0
Multiply (-1) on the above equation we get:
3x2+3y2−20x+32=03{{x}^{2}}+3{{y}^{2}}-20x+32=0
We have to divide this equation by 3 we get:
x2+y2−203x+323=0{{x}^{2}}+{{y}^{2}}-\dfrac{20}{3}x+\dfrac{32}{3}=0
As in the question they have mentioned the circle so, that we have this above equation in the form of circle equation that is to get that equation we need to add or subtract by 1009\dfrac{100}{9}
x2+y2−2×103x+1009+323−1009=0{{x}^{2}}+{{y}^{2}}-\dfrac{2\times 10}{3}x+\dfrac{100}{9}+\dfrac{32}{3}-\dfrac{100}{9}=0
By rearranging the term we get:
x2−2×103x+1009+y2+323−1009=0{{x}^{2}}-\dfrac{2\times 10}{3}x+\dfrac{100}{9}+{{y}^{2}}+\dfrac{32}{3}-\dfrac{100}{9}=0
By simplifying further we get:
(x−103)2+y2+323−1009=0{{\left( x-\dfrac{10}{3} \right)}^{2}}+{{y}^{2}}+\dfrac{32}{3}-\dfrac{100}{9}=0
By further solving we get:
⇒(x−103)2+y2+−49=0\Rightarrow {{\left( x-\dfrac{10}{3} \right)}^{2}}+{{y}^{2}}+\dfrac{-4}{9}=0
⇒(x−103)2+y2=49\Rightarrow {{\left( x-\dfrac{10}{3} \right)}^{2}}+{{y}^{2}}=\dfrac{4}{9}
If you notice this above equation is in the form of (x−h)2+(y−k)2=r2{{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}
⇒(x−103)2+(y−0)2=49\Rightarrow {{\left( x-\dfrac{10}{3} \right)}^{2}}+{{\left( y-0 \right)}^{2}}=\dfrac{4}{9}
⇒(x−103)2+(y−0)2=(23)2\Rightarrow {{\left( x-\dfrac{10}{3} \right)}^{2}}+{{\left( y-0 \right)}^{2}}=\left(\dfrac{2}{3}\right)^{2}
This represents a circle with center (103,0)\left( \dfrac{10}{3},0 \right) and radius 23\dfrac{2}{3}
So, the correct option is “option 4”.

Note: In this particular problem, we need to keep in mind the equation of the circle. For such types of problems, always make the equation in the form of a circle. Don’t make silly mistakes while simplifying the problems to avoid confusion, solve the problems step wise step. So, the above solution is preferred for such types of problems.