|z/(z+i)|=2 Represents a Circle — Find Distance of Centre From D(1,5)

JEE Main 2025. Previous Year Question.

Problem

If $\left|\dfrac{z}{z+i}\right|=2$ represents a circle with centre $P$, then the distance of $P$ from the point $D(1,5)$ is?

Note: the worked solution embedded below stops at $\dfrac{370}{9}$, which is the square of the distance. The question asks for the distance itself, so the answer here takes the final square root.

Key insight. A modulus ratio equation like $|z|=2|z+i|$ becomes an ordinary circle equation once $z=x+iy$ is substituted and both sides are squared — the trick is recognising the general form $x^2+y^2+2gx+2fy+c=0$ hiding inside after expansion, from which the centre reads off directly as $(-g,-f)$.

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Approach

Rewrite the equation as $|z|=2|z+i|$, substitute $z=x+iy$, and square both sides to remove the moduli. Expand and simplify into the general circle form to find the centre $P$. Then apply the distance formula between $P$ and $D(1,5)$ — being careful to take the final square root, since the distance formula’s intermediate result is the squared distance.

Solution

Step 1 — Rewrite the equation

$$\left|\frac{z}{z+i}\right|=2 \implies |z| = 2|z+i|$$

Step 2 — Substitute z = x + iy

$$|x+iy| = 2|x+i(y+1)|$$

Step 3 — Square both sides

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

Step 4 — Expand and simplify

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

$$0 = 3x^2+3y^2+8y+4$$

Dividing by 3:

$$x^2+y^2+\frac{8}{3}y+\frac{4}{3}=0$$

Step 5 — Identify the centre

Comparing with $x^2+y^2+2gx+2fy+c=0$: $2g=0\implies g=0$; $2f=\dfrac{8}{3}\implies f=\dfrac{4}{3}$.

$$P = (-g,-f) = \left(0,\ -\frac{4}{3}\right)$$

Step 6 — Compute the squared distance from P to D(1,5)

$$PD^2 = (0-1)^2+\left(-\frac{4}{3}-5\right)^2 = 1+\left(-\frac{19}{3}\right)^2 = 1+\frac{361}{9} = \frac{9+361}{9} = \frac{370}{9}$$

Step 7 — Take the square root to find the actual distance

$$PD = \sqrt{\frac{370}{9}} = \frac{\sqrt{370}}{3}$$

Answer

Common mistakes

• Stopping at the squared distance and reporting it as the final answer. $\dfrac{370}{9}$ is $PD^2$, not $PD$ — the question asks specifically for the distance, so the final square root in Step 7 isn’t optional. (If the original question instead asked for the square of the distance, $\dfrac{370}{9}$ would be the correct figure — but as stated here, it asks for the distance itself.)
• Getting the sign wrong when reading off the centre from the general circle form. The centre is $(-g,-f)$ — the negatives of the half-coefficients — not $(g,f)$ directly.

Practise next

• If $\left|\dfrac{z}{z-i}\right|=3$ represents a circle with centre $P$, find the distance of $P$ from the point $E(2,4)$, using the same substitute-and-expand method.