Question

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

Answer 1

Given that: ${\displaystyle \left|\frac{z-2}{z-3}\right|}$ = 2 

Let z = x + iy

∴ ${\displaystyle \left|\frac{x+iy-2}{x+iy-3}\right|}$ = 2

⇒ ${\displaystyle \left|\frac{(x-2)+iy}{(x-3)+iy}\right|}$ = 2  ......${\displaystyle [\because \left|\frac{a}{b}\right|=\frac{\left|a\right|}{\left|b\right|}]}$

⇒ ${\displaystyle |(x-2)+iy|=2|(x-3)+iy|}$

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

Squaring both sides, we get

(x – 2)² + y² = 4[(x – 3)² + y²]

⇒ x² + 4 – 4x + y² = 4[x² + 9 – 6x + y²]

⇒ x² + y² – 4x + 4 = 4x² + 4y² – 24x + 36

⇒ 3x² + 3y² – 20x + 32 = 0

⇒ ${\displaystyle x^2+y^2-\frac{20}{3}x+\frac{32}{3}}$ = 0

Here g = ${\displaystyle \frac{-10}{3}}$, f = 0

r = ${\displaystyle \sqrt{g^2+f^2-\text{c}}}$

= ${\displaystyle \sqrt{\frac{100}{9}+0-\frac{32}{3}}}$

= ${\displaystyle \sqrt{\frac{100}{9}-\frac{32}{3}}}$

= ${\displaystyle \sqrt{\frac{100-96}{9}}}$

= ${\displaystyle \sqrt{\frac{4}{9}}}$

= ${\displaystyle \frac{2}{3}}$

Hence, the required equation of the circle is ${\displaystyle x^2+y^2-\frac{20}{3}x+\frac{32}{3}}$ = 0

Centre = (–g, –f) = ${\displaystyle (\frac{10}{3},0)}$ and r = ${\displaystyle \frac{2}{3}}$.