Question

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

Or

If arg (z - 1) = arg (z + 3 i), find (x - 1) : y, where z = x + iy.

Answer 1

Let z = x + iy

Given equation is $\left|\frac{z-2}{z-3}\right|=2\Rightarrow \left|\frac{x+iy-2}{x+iy-3}\right|=2$

$\Rightarrow |x+iy-2|=2|x+iy-3|$

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

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

$\Rightarrow (x-2{)}^2+y^2=4[(x-3{)}^2+y^2]$ [squaring both sides]

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

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

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

This represents a equation of circle and its centre is $(\frac{10}{3},0)$

and radius $=\sqrt{{\left(\frac{10}{3}\right)}^2-\frac{32}{3}}=\sqrt{\frac{100}{9}-\frac{32}{3}}$

$[\because r=\sqrt{g^2+f^2-c}]$

$=\sqrt{\frac{100-96}{9}}=\frac{2}{3}\text{units}$

Or

Given that, arg (z-1) = arg (z + 3i)

$\Rightarrow$ arg (x + iy - 1) = arg (x + iy + 3i) [$\because$ z = x + iy]

$\Rightarrow$ arg (x - 1 + iy) = arg [x + i(y + 3)]

$\Rightarrow ta{n}^{-1}\left(\frac{y}{x-1}\right)=ta{n}^{-1}\left(\frac{y+3}{x}\right)$

$[\because arg(x+iy)=ta{n}^{-1}\frac{y}{x}]$

$\Rightarrow \frac{y}{(x-1)}=\frac{(y+3)}{x}\Rightarrow xy=(x-1)(y+3)$

$\Rightarrow xy=xy+3x-y-3\Rightarrow 3x-3=y$

$\Rightarrow \frac{x-1}{y}=\frac{1}{3}$

$\therefore x-1:y=1:3$