Given, (2−i)z=(2+i)z Let z=x+iy, then z=x−iy ⇒‌‌(2−i)(x+iy)=(2+i)(x−iy) ⇒2x−ix+2iy+y=2x+ix−2iy+y ⇒‌‌2ix−4iy=0 ∴ Equation of line L1⇒x−2y=0. . . (i) Also, (2+i)z+(i−2)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 ∴ Equation of line L2⇒x+2y−2=0. . . (ii) From Eqs. (i) and (ii), 4y=2‌ or ‌y=1∕2‌ and ‌x=1 Hence, centre =(1,1∕2) Equation of third line L3⇒iz+z+1+i=0 ⇒‌‌i(x+iy)+(x−iy)+1+i=0 ⇒‌‌ix−y+x−iy+1+i=0 ⇒‌‌(x−y+1)+i(x−y+1)=0 ∴ Radius = Distance of point (1,1∕2) to the line x−y+1=0 ∴‌‌r=‌