JEE Main 25 Feb 2021 Shift 1 Solved Paper

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=‌

|1− 1 2 +1|

√12+12

=‌

3

2√2