Let the lines (2 $-$ i)z = (2 + i)$\overline z$ and (2 $+$ i)z + (i $-$ 2)$\overline z$ $-$ 4i = 0, (here i² = $-$1) be normal to a circle C. If the line iz + $\overline z$ + 1 + i = 0 is tangent to this circle C, then its radius is :

$(2 - i)z = (2 + i)\overline z$<br><br>$\Rightarrow (2 - i)(x + iy) = (2 + i)(x - iy)$<br><br>$\Rightarrow 2x - ix + 2iy + y = 2x + ix - 2 - iy + y$<br><br>$\Rightarrow 2ix - 4iy = 0$<br><br>${L_1}:x - 2y = 0$<br><br>$\Rightarrow (2 + i)z + (i - 2)\overline z - 4i = 0$<br><br>$\Rightarrow (2 + i)(x + iy) + (i - 2)(x - iy) - 4i = 0$<br><br>$\Rightarrow 2x + ix + 2iy - y + ix - 2x + y + 2iy - 4i = 0$<br><br>$\Rightarrow 2ix + 4iy - 4i = 0$<br><br>${L_2}:x + 2y - 2 = 0$<br><br>Solve L₁ and L₂: x = 1, $4y = 2,y = {1 \over 2}$<br><br>$\therefore$ $x = 1$<br><br>Centre $\left( {1,{1 \over 2}} \right)$<br><br>${L_3}:iz + \overline z + 1 + i = 0$<br><br>$\Rightarrow i(x + iy) + x - iy + 1 + i = 0$<br><br>$\Rightarrow ix - y + x - iy + 1 + i = 0$<br><br>$\Rightarrow (x - y + 1) + i(x - y + 1) = 0$<br><br>Radius = distance from $\left( {1,{1 \over 2}} \right)$ to $x - y + 1 = 0$<br><br>$\Rightarrow$ $r = {{1 - {1 \over 2} + 1} \over {\sqrt 2 }}$<br><br>$\Rightarrow$ $r = {3 \over {2\sqrt 2 }}$