.Let `z` be a complex number such that `|(z-2i)/(z+i)|=2,z!=-i`.Then `z` lies on the circle of radius `2` and centre

Options

A. `(2,0)`
B. `(0,2)`
C. `(0,0)`
D. `(0,-2)`

Answer

To solve the problem, we need to analyze the given equation involving the complex number \( z \). The equation is

\[ \left| \frac{z - 2i}{z + i} \right| = 2 \]

where \( z \neq -i \). We will express \( z \) in terms of its real and imaginary parts and then manipulate the equation to find the center and radius of the circle on which \( z \) lies.

### Step 1: Express \( z \) in terms of real and imaginary parts
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers. Then we can rewrite the equation as:

\[ \left| \frac{(x + iy) - 2i}{(x + iy) + i} \right| = 2 \]

This simplifies to:

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

### Step 2: Use the property of modulus
Using the property of modulus of complex numbers, we have:

\[ \frac{|x + i(y - 2)|}{|x + i(y + 1)|} = 2 \]

### Step 3: Calculate the moduli
Now, we calculate the moduli:

\[ |x + i(y - 2)| = \sqrt{x^2 + (y - 2)^2} \]

\[ |x + i(y + 1)| = \sqrt{x^2 + (y + 1)^2} \]

Substituting these into our equation gives:

\[ \frac{\sqrt{x^2 + (y - 2)^2}}{\sqrt{x^2 + (y + 1)^2}} = 2 \]

### Step 4: Square both sides to eliminate the square roots
Squaring both sides results in:

\[ \frac{x^2 + (y - 2)^2}{x^2 + (y + 1)^2} = 4 \]

### Step 5: Cross-multiply to simplify
Cross-multiplying yields:

\[ x^2 + (y - 2)^2 = 4(x^2 + (y + 1)^2) \]

### Step 6: Expand both sides
Expanding both sides gives:

\[ x^2 + (y^2 - 4y + 4) = 4x^2 + 4(y^2 + 2y + 1) \]

This simplifies to:

\[ x^2 + y^2 - 4y + 4 = 4x^2 + 4y^2 + 8y + 4 \]

### Step 7: Rearranging the equation
Rearranging terms results in:

\[ 0 = 3x^2 + 3y^2 + 12y \]

Dividing through by 3 gives:

\[ 0 = x^2 + y^2 + 4y \]

### Step 8: Completing the square
We can complete the square for the \( y \) terms:

\[ x^2 + (y^2 + 4y + 4) = 4 \]

This can be rewritten as:

\[ x^2 + (y + 2)^2 = 4 \]

### Conclusion: Identify the center and radius
This is the equation of a circle with center at \( (0, -2) \) and radius \( 2 \).

Thus, the center of the circle is \( (0, -2) \) and the radius is \( 2 \).