Let the lines $(2-i)z=(2+i)\bar{z}$ and $(2+i)z+(i-2)\bar{z}-4i=0$, (here $i^{2}=-1$) be normal to a circle $C$. If the line $iz+\bar{z}+1+i=0$ is tangent to this circle $C$, then its radius is :

JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Complex Numbers

Let the lines

(2 - i) z = (2 + i) \overset{z}{ˉ}

and

(2 + i) z + (i - 2) \overset{z}{ˉ} - 4 i = 0

, (here

i^{2} = - 1

) be normal to a circle

C

. If the line

i z + \overset{z}{ˉ} + 1 + i = 0

is tangent to this circle

C

, then its radius is :

Answer: (B)

With

z = x + i y

, both sides of

(2 - i) z = (2 + i) \overset{z}{ˉ}

have real part

2 x + y

, and equating imaginary parts gives

2 y - x = x - 2 y

, that is

x - 2 y = 0

. The second line,

(2 + i) z + (i - 2) \overset{z}{ˉ} - 4 i = 0

, reduces to

i (2 x + 4 y - 4) = 0

, that is

x + 2 y = 2

. Both lines are normal to the circle, so they meet at its centre, and

x = 2 y

with

x + 2 y = 2

gives the centre

(1, \frac{1}{2})

. The tangent

i z + \overset{z}{ˉ} + 1 + i = 0

becomes

(x - y + 1) + i (x - y + 1) = 0

, that is

x - y + 1 = 0

, so the radius is the distance from the centre to this line,

r = \frac{1 - \frac{1}{2} + 1}{1 ^{2} + ( - 1 ) ^{2}} = \frac{3}{2 2} .

The radius of the circle is

\frac{3}{2 2}

JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Single correct option