Test Index

Complex Numbers Part 2

Section: Mathematics

© examsnet.com

Question : 37 of 106

Marks: +1, -0

Let

\left|\frac{\overline{z}-i}{2\overline{z}+i}\right|=\frac{1}{3},\ z\in C

, be the equation of a circle with center at

C

. If the area of the triangle, whose vertices are at the points

(0,0), C

(\alpha, 0)

is 11 square units, then

\alpha^2

[23 Jan 2025 Shift 1]

$\frac{81}{25}$
50
100
$\frac{121}{25}$

Solution:

Let

z=x+iy

, so

\overline{z}=x-iy

.

Then the given equation becomes

\left|\frac{x-i(y+1)}{2x+i(1-2y)}\right|=\frac{1}{3}.

So,

3\sqrt{x^2+(y+1)^2}=\sqrt{(2x)^2+(1-2y)^2}.

Squaring both sides:

9(x^2+y^2+2y+1)=4x^2+4y^2-4y+1.

Simplifying:

5x^2+5y^2+22y+8=0.

Thus, the center of the circle is

C\left(0,-\frac{11}{5}\right)

.

The triangle has vertices

(0,0)

,

C

, and

(\alpha,0)

.

Its base is

|\alpha|

and its height is

\left|-\frac{11}{5}\right|=\frac{11}{5}

.

Area of

\Delta

:

=\frac{1}{2}|\alpha|\left|-\frac{11}{5}\right|=11.

Thus,

\frac{11}{10}|\alpha|=11.

So,

|\alpha|=10

.

Therefore,

\alpha^2=100

.

Hence, the correct option is

\boxed{100}

.

© examsnet.com

Go to Question:

Prev Question

Next Question