Circle x²+y²=8x Meets Hyperbola x²/9−y²/4=1 at A,B — Centroid of PAB

JEE Main 2025 — 28 January, Evening Shift. Previous Year Question.

Problem

If $A$ and $B$ are the points of intersection of the circle $x^2+y^2-8x=0$ and the hyperbola $\dfrac{x^2}{9}-\dfrac{y^2}{4}=1$, and a point $P$ moves on the line $2x-3y+4=0$, then the centroid of $\triangle PAB$ lies on the line:

Key insight. Since $A$ and $B$ are fixed points and only $P$ moves, the centroid’s coordinates are just $A$, $B$’s fixed coordinates plus a moving point’s coordinates — divided by 3. Writing $P$ generically using the line it moves on turns the centroid into a parametrised point, and eliminating that parameter reveals the line it traces.

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Approach

First solve the circle and hyperbola equations simultaneously to pin down $A$ and $B$ exactly. Then write a generic point $P$ on the given line using one parameter, form the centroid formula symbolically, and eliminate the parameter to get a relation purely between the centroid’s $x$ and $y$ coordinates — that relation is the required line.

Solution

Step 1 — Eliminate y² between the two curves

From the hyperbola: $y^2 = \dfrac{4x^2}{9}-4$. Substituting into the circle $x^2+y^2-8x=0$:

$$x^2 + \frac{4x^2}{9}-4-8x = 0$$

Multiplying through by $9$:

$$9x^2+4x^2-36-72x=0 \implies 13x^2-72x-36=0$$

Step 2 — Solve for x

Factoring: $13x^2-78x+6x-36=0 \implies 13x(x-6)+6(x-6)=0 \implies (x-6)(13x+6)=0$

So $x=6$ or $x=-\dfrac{6}{13}$.

Step 3 — Reject the invalid root

At $x=-\dfrac{6}{13}$: $y^2 = \dfrac{4x^2}{9}-4$ works out to a negative value (since $x^2$ is tiny here), which is impossible. So this root is rejected, leaving $x=6$.

Step 4 — Find A and B

At $x=6$: $y^2 = \dfrac{4(36)}{9}-4 = 16-4=12 \implies y=\pm2\sqrt3$

So $A=(6,2\sqrt3)$ and $B=(6,-2\sqrt3)$.

Step 5 — Write a generic point P on the line

From $2x-3y+4=0$: $y = \dfrac{2x+4}{3}$. Let $P = \left(h,\ \dfrac{2h+4}{3}\right)$.

Step 6 — Write the centroid’s coordinates

$$X = \frac{6+6+h}{3} = \frac{12+h}{3}, \qquad Y = \frac{2\sqrt3-2\sqrt3+\frac{2h+4}{3}}{3} = \frac{2h+4}{9}$$

(The $y$-coordinates of $A$ and $B$ cancel since they’re equal and opposite.)

Step 7 — Eliminate h

From $X$: $h = 3X-12$. Substituting into $Y$:

$$Y = \frac{2(3X-12)+4}{9} = \frac{6X-20}{9} \implies 9Y = 6X-20$$

$$6X-9Y = 20$$

Answer

Common mistakes

• Forgetting to reject the negative-$y^2$ root. The quadratic in $x$ has two roots, but only one gives real values of $y$ — skipping this check risks carrying a spurious third “intersection point” into the rest of the problem.
• Not noticing the $y$-coordinates of $A$ and $B$ cancel. Since $A$ and $B$ share the same $x$-coordinate with opposite $y$-values, the centroid’s $y$-coordinate simplifies considerably — missing this makes the elimination step much messier than necessary.

Practise next

• If $A$ and $B$ are the intersection points of $x^2+y^2=25$ and $\dfrac{x^2}{8}-\dfrac{y^2}{9}=1$ that lie to the right of the $y$-axis, and $P$ moves on $x-2y+3=0$, find the line on which the centroid of $\triangle PAB$ lies, using the same elimination method.