Let the Circle C₁: x² + y² − 2(x + y) + 1 = 0 and C₂ Be a Circle With Centre (−1, 0) and Radius 2. If the Common Chord of C₁ and C₂ Intersects the y-axis at P, Then the Square of the Distance of P From the Centre of C₁ Is

JEE Main 2024 — 5 April, Evening Shift. Previous Year Question.

Problem

Let the circle $C_1: x^2+y^2-2(x+y)+1=0$ and $C_2$ be a circle having centre at $(-1,0)$ and radius $2$. If the line of the common chord of $C_1$ and $C_2$ intersects the $y$-axis at the point $P$, then the square of the distance of $P$ from the centre of $C_1$ is:

Key insight. The common chord of two circles is exactly their radical axis — subtracting one circle’s equation from the other (with both written in the same $x^2+y^2+\ldots$ form) makes the $x^2$ and $y^2$ terms cancel, leaving a straight line directly.

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Approach

Write $C_2$’s equation explicitly from its given centre and radius. Subtract $C_1$’s equation from $C_2$’s (or vice versa) to get the common chord’s equation — the radical axis — since both circles share the same $x^2+y^2$ terms. Find where this line crosses the $y$-axis to locate $P$, then use the distance formula between $P$ and $C_1$’s centre.

Solution

Step 1 — Write C₁ in standard form

$$C_1: x^2+y^2-2x-2y+1=0$$

Step 2 — Write C₂’s equation from its centre and radius

Centre $(-1,0)$, radius $2$: $(x+1)^2+y^2=4$

$$x^2+2x+1+y^2=4 \implies x^2+y^2+2x-3=0$$

Step 3 — Find the common chord using S₁ − S₂ = 0

$$\left(x^2+y^2-2x-2y+1\right)-\left(x^2+y^2+2x-3\right) = 0$$

$$-4x-2y+4=0$$

Dividing by $-2$:

$$2x+y-2=0$$

Step 4 — Find where this line meets the y-axis

Setting $x=0$: $y=2$. So $P=(0,2)$.

Step 5 — Find the centre of C₁

Comparing $C_1$’s equation to $x^2+y^2+2gx+2fy+c=0$: $2g=-2\implies g=-1$; $2f=-2\implies f=-1$. Centre $=(-g,-f)=(1,1)$.

Step 6 — Compute PC₁²

$$PC_1^2 = (1-0)^2+(1-2)^2 = 1+1 = 2$$

Answer

Common mistakes

• Subtracting the two circle equations in the wrong order, or before both are in the same standard form. $S_1-S_2$ only cancels the $x^2$ and $y^2$ terms cleanly if both circles are written as $x^2+y^2+\ldots=0$ with matching leading coefficients — converting $C_2$’s centre-radius form into this exact shape first is essential.
• Confusing the centre of $C_1$ with the centre of $C_2$ when computing the final distance. The question specifically asks for the distance from $P$ to $C_1$’s centre, not $C_2$’s (which was given directly as $(-1,0)$ and is easy to reach for by mistake).

Practise next

• Let $C_1: x^2+y^2-4x-2y+1=0$ and $C_2$ have centre $(1,-1)$ and radius $3$. If their common chord meets the $x$-axis at $Q$, find the square of the distance of $Q$ from the centre of $C_1$, using the same radical-axis approach.