A Line Segment AB of Length λ Moves Such That A and B Remain on the Periphery of a Circle of Radius λ. Then the Locus of the Point That Divides AB in the Ratio 2:3 Is a Circle of Radius

JEE Main 2023 — 10 April, Morning Shift. Previous Year Question.

Problem

A line segment $AB$ of length $\lambda$ moves such that the points $A$ and $B$ remain on the periphery of a circle of radius $\lambda$. Then the locus of the point that divides the line segment $AB$ in the ratio $2:3$, is a circle of radius:

Key insight. Since chord $AB$ has the same length as the circle’s own radius, the angle it subtends at the centre is fixed at exactly $60°$, no matter where the chord sits on the circle. That fixed angle is what makes the dividing point’s locus a genuine circle rather than some more complicated curve.

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Approach

Show that a chord of length $\lambda$ on a circle of radius $\lambda$ subtends a fixed angle of $60°$ at the centre. Parametrise $A$ and $B$ using angles $\theta_1,\theta_2$ with $\theta_1-\theta_2=60°$, find the point $P$ dividing $AB$ in ratio $2:3$ using the section formula, and compute $|OP|^2$ — it should come out independent of $\theta_1,\theta_2$ individually, confirming $P$ traces a circle.

Solution

Step 1 — Show the chord subtends 60° at the centre

With $A=\lambda(\cos\theta_1,\sin\theta_1)$ and $B=\lambda(\cos\theta_2,\sin\theta_2)$ on the circle:

$$|AB|^2 = \lambda^2\left[(\cos\theta_1-\cos\theta_2)^2+(\sin\theta_1-\sin\theta_2)^2\right] = \lambda^2\left[2-2\cos(\theta_1-\theta_2)\right]$$

Setting $|AB|=\lambda$: $\lambda^2 = 2\lambda^2\left[1-\cos(\theta_1-\theta_2)\right] \implies \cos(\theta_1-\theta_2)=\dfrac12 \implies \theta_1-\theta_2=60°$

Step 2 — Find point P dividing AB in ratio 2:3

Using the section formula (dividing internally in ratio $2:3$ from $A$ to $B$):

$$P = \frac{3A+2B}{5}$$

$$P_x = \frac{\lambda}{5}\left(3\cos\theta_1+2\cos\theta_2\right), \qquad P_y = \frac{\lambda}{5}\left(3\sin\theta_1+2\sin\theta_2\right)$$

Step 3 — Compute |OP|² (with O the circle’s centre)

$$|OP|^2 = \frac{\lambda^2}{25}\left[(3\cos\theta_1+2\cos\theta_2)^2+(3\sin\theta_1+2\sin\theta_2)^2\right]$$

Step 4 — Expand

$$= \frac{\lambda^2}{25}\left[9\cos^2\theta_1+12\cos\theta_1\cos\theta_2+4\cos^2\theta_2+9\sin^2\theta_1+12\sin\theta_1\sin\theta_2+4\sin^2\theta_2\right]$$

$$= \frac{\lambda^2}{25}\left[9+4+12\left(\cos\theta_1\cos\theta_2+\sin\theta_1\sin\theta_2\right)\right]$$

$$= \frac{\lambda^2}{25}\left[13+12\cos(\theta_1-\theta_2)\right]$$

Step 5 — Substitute cos(θ₁−θ₂) = 1/2

$$|OP|^2 = \frac{\lambda^2}{25}\left[13+12\times\frac12\right] = \frac{\lambda^2}{25}(13+6) = \frac{19\lambda^2}{25}$$

Step 6 — Take the square root

Since $|OP|^2$ is independent of $\theta_1,\theta_2$ individually (only their difference matters, which is fixed at $60°$), $P$ traces a circle centred at $O$ with:

$$|OP| = \sqrt{\frac{19\lambda^2}{25}} = \frac{\sqrt{19}}{5}\lambda$$

Answer

Common mistakes

• Assuming the subtended angle changes as the chord moves around the circle. Since both the chord length and the radius are fixed at $\lambda$, the angle they determine is a fixed geometric relationship — the chord can slide anywhere on the circle, but the subtended angle stays exactly $60°$ throughout.
• Stopping at $|OP|^2$ and forgetting the final square root. The computed quantity $\dfrac{19\lambda^2}{25}$ is the square of the locus radius — the actual radius is $\dfrac{\sqrt{19}}{5}\lambda$, and mismatching this against the answer options (which are given as un-squared radii) would lead to picking the wrong option.

Practise next

• A line segment $PQ$ of length $\lambda$ moves such that $P$ and $Q$ remain on a circle of radius $\lambda$. Find the radius of the locus of the point dividing $PQ$ in ratio $1:4$, using the same fixed-subtended-angle approach.