Let the lines (2-i) z=(2+i) {z} and (2+i) z+(i-2) {z}-4 i=0, (here i²=-1 ) be normal to a circle C. If the line i z+{z}+1+i=0 is tangent to this circle C, then its radius is [25 Feb 2021 Shift 1]

Correct Answer is : 322\;{3}{2{2}}223

Correct Answer is : 322\;{3}{2{2}}223

Test Index

Complex Numbers Part 1

Section: Mathematics

Question : 33 of 100

Marks: +1, -0

Let the lines

(2-i) z=(2+i) \overline{z}

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

i^{2}=-1

C

i z+\overline{z}+1+i=0

C

Solution:

Given,

(2-i) z=(2+i) \overline{z}

Let

z=x+i y

, then

\overline{z}=x-i y

\Rightarrow \;\; (2-i)(x+i y)=(2+i)(x-i y)

\Rightarrow 2 x-i x+2 i y+y=2 x+i x-2 i y+y

\Rightarrow \;\; 2 i x-4 i y=0

\therefore

Equation of line

L_{1} \Rightarrow x-2 y=0

. . . (i)

Also,

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

\Rightarrow \;\; (2+i)(x+i y)+(i-2)(x-i y)-4 i=0

\Rightarrow \;\; 2 x+i x+2 i y-y+i x-2 x+y+2 i y-4 i=0

\Rightarrow \;\; 2 i x+4 i y-4 i=0

\therefore

Equation of line

L_{2} \Rightarrow x+2 y-2=0

. . . (ii)

From Eqs. (i) and (ii),

4 y=2 \; \text{ or } \; y=\frac{1}{2} \; \text{ and } \; x=1

Hence, centre

=\left(1,\frac{1}{2}\right)

Equation of third line

L_{3} \Rightarrow i z+\overline{z}+1+i=0

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

\Rightarrow \;\; i x-y+x-i y+1+i=0

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

\therefore

Radius

=

Distance of point

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

to the line

x-y+1=0

\therefore \;\; r=\frac{\left|1-\frac{1}{2}+1\right|}{\sqrt{1^{2}+1^{2}}}=\frac{3}{2\sqrt{2}}

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