Let the lines (2 - i)z = (2 + i) ¯ z and (2 + i)z + (i - 2) ¯ z -...

$( 2 - i ) z = ( 2 + i ) \overset{¯}{z}$

$⇒ ( 2 - i ) ( x + iy ) = ( 2 + i ) ( x - iy )$

$⇒ 2x - ix + 2iy + y = 2x + ix - 2iy + y$

$⇒ 2ix - 4iy = 0$

${L}_{1}: x - 2y = 0$

$⇒ ( 2 + i ) z + ( i - 2 ) \overset{¯}{z} - 4i = 0$

$⇒ ( 2 + i ) ( x + iy ) + ( i - 2 ) ( x - iy ) - 4i = 0$

.$⇒ 2x + ix + 2iy - y + ix - 2x + y + 2iy - 4i = 0$

$⇒ 2ix + 4iy - 4i = 0$

Solve ${L}_{1}$ and ${L}_{2}, 4y = 2, y = \dfrac{1}{2}$

$∴ x = 1$

Centre \left(1, \dfrac{1}{2}\right)$</p><p>${L}_{3}: iz + \overset{¯}{z} + 1 + i = 0$</p><p>$⇒ i ( x + iy ) + x - iy + 1 + i = 0$</p><p>$⇒ ix - y + x - iy + 1 + i = 0$</p><p>$⇒ ( x - y + 1 ) + i ( x - y + 1 ) = 0$</p><p>Radius = distance from \left(1, \dfrac{1}{2}\right)$ to $x - y + 1 = 0$

$r = \left|\dfrac{1 - \dfrac{1}{2} + 1}{\sqrt{2}}\right|$

$r = \dfrac{3}{2\sqrt{2}}$