JEE Mathematics 2023 Question with Solution

Given: $\dfrac{|z - 2i|}{|z + i|} = 2$ and $z \neq -i$.

Find: The centre of the circle on which $z$ lies.

Let $z = x + yi$, where $x, y \in \mathbb{R}$. Then

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

Squaring both sides,

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

Cross-multiplying,

$$x^2 + (y - 2)^2 = 4\left(x^2 + (y + 1)^2\right)$$

Expanding,

$$x^2 + y^2 - 4y + 4 = 4x^2 + 4y^2 + 8y + 4$$

Bringing all terms to one side,

$$-3x^2 - 3y^2 - 12y = 0$$

So,

$$3x^2 + 3y^2 + 12y = 0$$ $$x^2 + y^2 + 4y = 0$$

Completing the square in $y$,

$$x^2 + (y^2 + 4y + 4) = 4$$ $$x^2 + (y + 2)^2 = 4$$

This is a circle with centre $(0, -2)$ and radius $2$.

Therefore, the correct option is D.

Given: $\dfrac{|z - 2i|}{|z + i|} = 2$.

Find: The centre of the corresponding circle.

Using modulus squared,

$$|z - 2i|^2 = 4|z + i|^2$$

That is,

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

On simplification, this gives the Cartesian form

$$x^2 + y^2 + 4y = 0$$

where $z = x + yi$.

Now complete the square:

$$x^2 + y^2 + 4y = 0$$ $$x^2 + (y + 2)^2 = 4$$

Hence the locus is a circle of radius $2$ with centre $(0, -2)$.

Therefore, the correct option is D.