If z=x+iy is a complex number satisfying left | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the equation: $\left|\frac{z-2i}{z+2i}\right|=2$.
Substituting $z=x+iy$: $\left|\frac{x+i(y-2)}{x+i(y+2)} \right|=2$.
Squaring both sides:

Vedclass · Accepted Answer

$\frac{8}{3}$

Vedclass · Answer

(C) Given the equation: $\left|\frac{z-2i}{z+2i}\right|=2$.
Substituting $z=x+iy$: $\left|\frac{x+i(y-2)}{x+i(y+2)} \right|=2$.
Squaring both sides: $\frac{x^2+(y-2)^2}{x^2+(y+2)^2}=4$.
$x^2+y^2-4y+4 = 4(x^2+y^2+4y+4)$.
$x^2+y^2-4y+4 = 4x^2+4y^2+16y+16$.
$3x^2+3y^2+20y+12=0$.
Dividing by $3$: $x^2+y^2+\frac{20}{3}y+4=0$.
Completing the square for $y$: $x^2+(y+\frac{10}{3})^2 = \frac{100}{9}-4 = \frac{100-36}{9} = \frac{64}{9}$.
Thus,$x^2+(y+\frac{10}{3})^2 = (\frac{8}{3})^2$.
The radius of the circle is $\frac{8}{3}$.

If $z=x+iy$ is a complex number satisfying $\left|\frac{z-2i}{z+2i}\right|=2$ and the locus of $z$ is a circle,then its radius is

Similar Questions

Let $z \neq -i$ be any complex number such that $\frac{z - i}{z + i}$ is a purely imaginary number. Then $z + \frac{1}{z}$ is

Let $z = x + iy$, where $x$ and $y$ are real. The points $(x, y)$ in the $X-Y$ plane for which $\frac{z+i}{z-i}$ is purely imaginary, lie on

If $S = \{z \in \mathbb{C} : |z - i| = |z + i| = |z - 1|\}$,then $n(S)$ is:

The maximum distance from the origin of coordinates to the point $z$ satisfying the equation $\left| z + \frac{1}{z} \right| = a$ is

If $z$ is a complex number,then the curves $|z|=1$,$|z-2|=1$,and $|z-1|=0$ have a common point at

Vedclass Test Series

Exam Paper Generator

Online Exam Module