Why is the centre and not from ?

The standard circle equation is . Here, , so .

JEE MainMathematics

Reviewed by official_key

+4 marks−1 if incorrectSingle correctpyq

Complex Numbers and Quadratic Equations: Mathematics | JEE Main

Q: Do internal and external division points always form the diameter?

Yes, for any ratio , the internal and external angle bisectors are perpendicular, making the segment connecting the two division points the diameter of the Apollonius circle.

Let z be a complex number such that

\left|\frac{z - 2i}{z + i}\right| = 2, z \neq -i

. Then z lies on the circle of radius 2 and centre

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

(2, 0)

(0, 2)

(0, -2)

(0, 0)

Step-by-step solutionView

Correct answer

By letting $z = x + iy$ or using the Apollonius circle property, the equation simplifies to $x^2 + (y + 2)^2 = 4$ , which is a circle with centre $(0, -2)$ and radius $2$ .

Option analysis

Why each option works or fails

A ·

(2, 0)

Confusing the real and imaginary components when reading the centre from the standard circle equation $x^2 + (y - k)^2 = r^2$ . Identify that the linear term appears in $y$ (the imaginary part), meaning the centre lies on the $y$ -axis with coordinates $(0, y_0)$ , not on the $x$ -axis.

B ·

(0, 2)

Making a sign error during the expansion of $|z - 2i|^2 = 4|z + i|^2$ , resulting in $y - 2$ instead of $y + 2$ in the completed square form. Carefully expand $(y - 2)^2 = y^2 - 4y + 4$ and $4(y + 1)^2 = 4y^2 + 8y + 4$ ; moving all terms to one side gives $3y^2 + 12y = 0$ , leading to $(y + 2)^2$ .

C ·

(0, -2)

Correct option: expanding $|x + i(y - 2)|^2 = 4|x + i(y + 1)|^2$ directly yields $x^2 + y^2 + 4y = 0$ , giving centre $(0, -2)$ . Ensure to keep track of the signs during the completion of the square: $y^2 + 4y = (y + 2)^2 - 4$ , which gives centre $(0, -2)$ .

D ·

(0, 0)

Assuming that the symmetry of the endpoints $2i$ and $-i$ balances out to the origin without expanding the squared modulus equation. Calculate the Apollonius division points explicitly: internal division gives $z = 0$ and external division gives $z = -4i$ , whose midpoint is $-2i$ , or $(0, -2)$ .

Reviewed route

Solution

StepWorking

01given

A complex number $z = x + iy$ satisfies $\left|\frac{z - 2i}{z + i}\right| = 2$ where $z \neq -i$ .

02goal

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

03approach

Substitute $z = x + iy$ , take the modulus on both sides, square both sides to eliminate square roots, and complete the square in Cartesian form.

04execute

Substitute $z = x + iy$ : $\left|\frac{x + i(y - 2)}{x + i(y + 1)}\right| = 2 \implies \frac{x^2 + (y - 2)^2}{x^2 + (y + 1)^2} = 4$ Expand both sides: $x^2 + y^2 - 4y + 4 = 4(x^2 + y^2 + 2y + 1)$ $x^2 + y^2 - 4y + 4 = 4x^2 + 4y^2 + 8y + 4$ Collect all terms on one side: $3x^2 + 3y^2 + 12y = 0 \implies x^2 + y^2 + 4y = 0$ Rewrite in standard circle form: $x^2 + (y + 2)^2 = 4 = 2^2$ Thus, the centre is $(0, -2)$ and radius is $2$ .

✓verify

The circle $x^2 + (y + 2)^2 = 4$ has centre $(0, -2)$ and radius $2$ . Note $z = -i$ gives $(0, -1)$ , which lies inside the circle since $0^2 + (-1+2)^2 = 1 < 4$ , so $z \neq -i$ is respected.

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why is the centre

(0, -2)

and not

(0, 2)

from

y^2 + 4y = 0

The standard circle equation is $(x - h)^2 + (y - k)^2 = r^2$ . Here, $y^2 + 4y = (y + 2)^2 - 4$ , so $k = -2$ .

Do internal and external division points always form the diameter?

Yes, for any ratio $k \neq 1$ , the internal and external angle bisectors are perpendicular, making the segment connecting the two division points the diameter of the Apollonius circle.

Answer

By letting $z = x + iy$ or using the Apollonius circle property, the equation simplifies to $x^2 + (y + 2)^2 = 4$ , which is a circle with centre $(0, -2)$ and radius $2$ .

Why each option works or fails

A: $(2, 0)$ - Confusing the real and imaginary components when reading the centre from the standard circle equation $x^2 + (y - k)^2 = r^2$ . Identify that the linear term appears in $y$ (the imaginary part), meaning the centre lies on the $y$ -axis with coordinates $(0, y_0)$ , not on the $x$ -axis.
B: $(0, 2)$ - Making a sign error during the expansion of $|z - 2i|^2 = 4|z + i|^2$ , resulting in $y - 2$ instead of $y + 2$ in the completed square form. Carefully expand $(y - 2)^2 = y^2 - 4y + 4$ and $4(y + 1)^2 = 4y^2 + 8y + 4$ ; moving all terms to one side gives $3y^2 + 12y = 0$ , leading to $(y + 2)^2$ .
C · correct: $(0, -2)$ - Correct option: expanding $|x + i(y - 2)|^2 = 4|x + i(y + 1)|^2$ directly yields $x^2 + y^2 + 4y = 0$ , giving centre $(0, -2)$ . Ensure to keep track of the signs during the completion of the square: $y^2 + 4y = (y + 2)^2 - 4$ , which gives centre $(0, -2)$ .
D: $(0, 0)$ - Assuming that the symmetry of the endpoints $2i$ and $-i$ balances out to the origin without expanding the squared modulus equation. Calculate the Apollonius division points explicitly: internal division gives $(0, -2)$ and external division gives $z = -4i$ , whose midpoint is $-2i$ , or $(0, -2)$ .

Step-by-step solution

given: A complex number $z = x + iy$ satisfies $\left|\frac{z - 2i}{z + i}\right| = 2$ where $z \neq -i$ .
goal: Find the centre of the circle on which $z$ lies.
approach: Substitute $z = x + iy$ , take the modulus on both sides, square both sides to eliminate square roots, and complete the square in Cartesian form.
execute: Substitute $z = x + iy$ : $\left|\frac{x + i(y - 2)}{x + i(y + 1)}\right| = 2 \implies \frac{x^2 + (y - 2)^2}{x^2 + (y + 1)^2} = 4$ Expand both sides: $x^2 + y^2 - 4y + 4 = 4(x^2 + y^2 + 2y + 1)$ $x^2 + y^2 - 4y + 4 = 4x^2 + 4y^2 + 8y + 4$ Collect all terms on one side: $3x^2 + 3y^2 + 12y = 0 \implies x^2 + y^2 + 4y = 0$ Rewrite in standard circle form: $x^2 + (y + 2)^2 = 4 = 2^2$ Thus, the centre is $(0, -2)$ and radius is $2$ .
verify: The circle $x^2 + (y + 2)^2 = 4$ has centre $(0, -2)$ and radius $2$ . Note $z = -i$ gives $(0, -1)$ , which lies inside the circle since $0^2 + (-1+2)^2 = 1 < 4$ , so $z \neq -i$ is respected.

Shortcut:

When to use it: Whenever locus is $|z - z_1| / |z - z_2| = k$ ( $k \neq 1$ ), using Apollonius circle diameter endpoints.

given: The points are $A(0, 2)$ and $B(0, -1)$ , and $\frac{PA}{PB} = 2$ .

goal: Find the centre of the circle directly via the internal and external division points of segment $AB$ .

approach: The locus of $P$ such that $PA/PB = 2$ is an Apollonius circle. Its diameter endpoints divide $AB$ internally and externally in the ratio $2:1$ .

execute: Internal point $P_1 = \frac{2(-i) + 1(2i)}{2 + 1} = \frac{0}{3} = 0 = (0, 0)$ . External point $P_2 = \frac{2(-i) - 1(2i)}{2 - 1} = -4i = (0, -4)$ . The centre is the midpoint of $P_1$ and $P_2$ : $\text{Centre} = \frac{0 + (-4i)}{2} = -2i = (0, -2).$

verify: Radius is $\frac{|P_1 - P_2|}{2} = \frac{4}{2} = 2$ , which matches the given radius.

Complex Numbers and Quadratic Equations: Mathematics | JEE Main

What feels right?

By letting $z = x + iy$ or using the Apollonius circle property, the equation simplifies to $x^2 + (y + 2)^2 = 4$ , which is a circle with centre $(0, -2)$ and radius $2$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

✓verify

We think you should solve this next

If $z$ is a complex number such that $|z| \ge 1$ , then the minimum value of…

If $\mathrm{z}=\mathrm{x}+\mathrm{i} y, \quad x y \neq 0$ , satisfies the equation…

Let $\alpha$ and $\beta$ be the roots of the equation…

Students also ask

Complex Numbers and Quadratic Equations: Mathematics | JEE Main

By letting z=x+iyz = x + iyz=x+iy or using the Apollonius circle property, the equation simplifies to x2+(y+2)2=4x^2 + (y + 2)^2 = 4x2+(y+2)2=4, which is a circle with centre (0,−2)(0, -2)(0,−2) and radius 222.

01given

02goal

03approach

04execute

✓verify

If zzz is a complex number such that ∣z∣≥1|z| \ge 1∣z∣≥1, then the minimum value of…

If z=x+iy,xy≠0\mathrm{z}=\mathrm{x}+\mathrm{i} y, \quad x y \neq 0z=x+iy,xy=0, satisfies the equation…

Let α\alphaα and β\betaβ be the roots of the equation…

Students also ask

By letting $z = x + iy$ or using the Apollonius circle property, the equation simplifies to $x^2 + (y + 2)^2 = 4$ , which is a circle with centre $(0, -2)$ and radius $2$ .

If $z$ is a complex number such that $|z| \ge 1$ , then the minimum value of…

If $\mathrm{z}=\mathrm{x}+\mathrm{i} y, \quad x y \neq 0$ , satisfies the equation…

Let $\alpha$ and $\beta$ be the roots of the equation…