Let the lines (2-i)z=(2+i) z and (2+i)z+(i-2) z-4i=0, (here i^2=-1) be normal to a circle C. If the line iz+ z+1+i=0 is tangent to this circle C,…
JEE Main 2021 — Mathematics Algebra
2021mcqhard
Let the lines (2−i)z=(2+i)zˉ and (2+i)z+(i−2)zˉ−4i=0, (here i2=−1) be normal to a circle C. If the line iz+zˉ+1+i=0 is tangent to this circle C, then its radius is :
Official previous-year question
Held on 25 Feb 2021 · Verified 6 Jul 2026.
Options
A
23
B
32
C
223
D
221
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Solution
(2−i)z=(2+i)zˉ
⇒(2−i)(x+iy)=(2+i)(x−iy)
⇒2x−ix+2iy+y=2x+ix−2iy+y
⇒2ix−4iy=0
L1:x−2y=0
⇒(2+i)z+(i−2)zˉ−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 L1 and L2,4y=2,y=21
∴x=1
Centre (1,21)
L3:iz+zˉ+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 from (1,21) to x−y+1=0
r=∣21−21+1∣
r=223
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