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QJEEM21Ma0e1c: JEE Main 2021 Mathematics - Class 11 JEEM21Ma0e1c Solution: Let the lines (2 i)z = (2 + i) and (2 i)z + (i 2) 4i = 0, (here …

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

Multiple Choice Options

  • A 322{3 \over {2\sqrt 2 }}
  • B 323\sqrt 2
  • C 122{1 \over {2\sqrt 2 }}
  • D 32{3 \over {\sqrt 2 }}

Step-by-Step Verified Solution

(2−i)z=(2+i)z‾(2 - i)z = (2 + i)\overline z

⇒(2−i)(x+iy)=(2+i)(x−iy)\Rightarrow (2 - i)(x + iy) = (2 + i)(x - iy)

⇒2x−ix+2iy+y=2x+ix−2−iy+y\Rightarrow 2x - ix + 2iy + y = 2x + ix - 2 - iy + y

⇒2ix−4iy=0\Rightarrow 2ix - 4iy = 0

L1:x−2y=0{L_1}:x - 2y = 0

⇒(2+i)z+(i−2)z‾−4i=0\Rightarrow (2 + i)z + (i - 2)\overline z - 4i = 0

⇒(2+i)(x+iy)+(i−2)(x−iy)−4i=0\Rightarrow (2 + i)(x + iy) + (i - 2)(x - iy) - 4i = 0

⇒2x+ix+2iy−y+ix−2x+y+2iy−4i=0\Rightarrow 2x + ix + 2iy - y + ix - 2x + y + 2iy - 4i = 0

⇒2ix+4iy−4i=0\Rightarrow 2ix + 4iy - 4i = 0

L2:x+2y−2=0{L_2}:x + 2y - 2 = 0

Solve L1 and L2: x = 1, 4y=2,y=124y = 2,y = {1 \over 2}

∴\therefore x=1x = 1

Centre (1,12)\left( {1,{1 \over 2}} \right)

L3:iz+z‾+1+i=0{L_3}:iz + \overline z + 1 + i = 0

⇒i(x+iy)+x−iy+1+i=0\Rightarrow i(x + iy) + x - iy + 1 + i = 0

⇒ix−y+x−iy+1+i=0\Rightarrow ix - y + x - iy + 1 + i = 0

⇒(x−y+1)+i(x−y+1)=0\Rightarrow (x - y + 1) + i(x - y + 1) = 0

Radius = distance from (1,12)\left( {1,{1 \over 2}} \right) to x−y+1=0x - y + 1 = 0

⇒\Rightarrow r=1−12+12r = {{1 - {1 \over 2} + 1} \over {\sqrt 2 }}

⇒\Rightarrow r=322r = {3 \over {2\sqrt 2 }}

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