If a Tangent to the Circle x² + y² = 1 Intersects the Coordinate Axes at Distinct Points P and Q, Then the Locus of the Midpoint of PQ Is

JEE Main 2019 — Previous Year Question.

Problem

If a tangent to the circle $x^2+y^2=1$ intersects the coordinate axes at distinct points $P$ and $Q$, then the locus of the midpoint of $PQ$ is:

Key insight. Writing the point of tangency in parametric form, $(\cos\theta,\sin\theta)$, gives the tangent line a single-parameter equation — this makes both axis intercepts, and hence the midpoint, explicit functions of $\theta$ alone, so eliminating $\theta$ via $\sin^2\theta+\cos^2\theta=1$ is all that’s needed to find the locus.

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Approach

Take the point of tangency on the unit circle as $(\cos\theta,\sin\theta)$, and write down the equation of the tangent there. Find where this tangent meets the $x$-axis (point $P$) and the $y$-axis (point $Q$), each in terms of $\theta$. Express the midpoint $(h,k)$ of $PQ$ in terms of $\cos\theta$ and $\sin\theta$, solve those two relations for $\cos\theta$ and $\sin\theta$ individually, and substitute into the Pythagorean identity to eliminate $\theta$ entirely.

Solution

Step 1 — Write the tangent at a parametric point

For the unit circle $x^2+y^2=1$, the tangent at $(\cos\theta,\sin\theta)$ is:

$$x\cos\theta+y\sin\theta = 1$$

Step 2 — Find P (intersection with the x-axis)

Setting $y=0$:

$$x=\frac1{\cos\theta} \implies P=\left(\frac1{\cos\theta},\,0\right)$$

Step 3 — Find Q (intersection with the y-axis)

Setting $x=0$:

$$y=\frac1{\sin\theta} \implies Q=\left(0,\,\frac1{\sin\theta}\right)$$

Step 4 — Write the midpoint of PQ

Let $(h,k)$ be the midpoint:

$$h = \frac{\frac1{\cos\theta}+0}2 = \frac1{2\cos\theta}, \qquad k = \frac{0+\frac1{\sin\theta}}2 = \frac1{2\sin\theta}$$

Step 5 — Solve for cos θ and sin θ

$$\cos\theta = \frac1{2h}, \qquad \sin\theta = \frac1{2k}$$

Step 6 — Eliminate θ using sin²θ + cos²θ = 1

$$\left(\frac1{2h}\right)^2+\left(\frac1{2k}\right)^2 = 1 \implies \frac1{4h^2}+\frac1{4k^2}=1$$

Multiplying through by $4h^2k^2$:

$$k^2+h^2 = 4h^2k^2$$

Step 7 — Rewrite using x and y

Substituting $h=x$, $k=y$:

$$x^2+y^2 = 4x^2y^2 \implies x^2+y^2-4x^2y^2=0$$

Answer

Common mistakes

• Working with an arbitrary point $(x_1,y_1)$ on the circle instead of the parametric form. This is possible but leaves an extra variable to eliminate at the end (since $(x_1,y_1)$ itself has to satisfy $x_1^2+y_1^2=1$) — using $(\cos\theta,\sin\theta)$ builds that constraint in from the start, via the Pythagorean identity, with nothing extra to substitute back.
• Mixing up which coordinate goes with which trigonometric function. $P$ (on the $x$-axis) comes from setting $y=0$ and involves $\cos\theta$; $Q$ (on the $y$-axis) comes from setting $x=0$ and involves $\sin\theta$ — swapping these swaps $h$ and $k$’s roles and can lead to comparing against the wrong answer choice.
• Forgetting to multiply through by $4h^2k^2$ before comparing with the options. Leaving the locus as $\dfrac1{4h^2}+\dfrac1{4k^2}=1$ is mathematically equivalent but not directly comparable with the polynomial-form answer choices.

Practise next

• If a tangent to the circle $x^2+y^2=4$ intersects the coordinate axes at distinct points $P$ and $Q$, find the locus of the midpoint of $PQ$.