Let C_1 and C_2 be the centres of the circles x²+y²-2x-2y-2 = 0 and x²+y²-6x-6y+14 = 0 respectively.

Q: Wait, why can't we just find the coordinates of $P$ and $Q$ and use the determinant formula for area?

You absolutely can! In coordinate geometry, there is often more than one path to the summit. However, in a high-stakes exam like JEE Advanced, time is your most precious currency. Solving for the intersection points involves solving a system of quadratic equations, which is prone to calculation errors. By recognizing the geometric properties—that this is a rhombus and specifically a square—you bypass the algebra entirely and arrive at the answer in seconds.

Q: How do I know the quadrilateral is a rhombus?

A rhombus is defined as a quadrilateral where all four sides are equal. In our case, the sides are $C_1P, C_1Q, C_2P,$ and $C_2Q$. Since $P$ and $Q$ lie on both circles, $C_1P$ and $C_1Q$ are radii of the first circle ($r_1 = 2$), and $C_2P$ and $C_2Q$ are radii of the second circle ($r_2 = 2$). Because $r_1 = r_2 = 2$, all four sides are equal to 2. Thus, it is a rhombus.

Q: Is it always a square? What if the radii were different?

Great question! It is not always a square. It is only a square when the distance between the centers $C_1C_2$ satisfies the condition $C_1C_2 = r\sqrt{2}$. If the radii were different or the distance between centers was different, the shape would remain a rhombus. In that general case, you would use the formula for the area of a rhombus: $\text{Area} = \frac{1}{2} \times d_1 \times d_2$, where $d_1$ and $d_2$ are the lengths of the diagonals (the common chord $PQ$ and the line of centers $C_1C_2$).

Question

Let $C_1$ and $C_2$ be the centres of the circles $x^2+y^2-2x-2y-2 = 0$ and $x^2+y^2-6x-6y+14 = 0$ respectively. If P and Q are the points of intersection of these circles, then the area (in sq. units) of the quadrilateral $PC_1QC_2$ is :

Options:

(A) 8

(B) 6

(C) 9

(D) 4

Sigma Percentile · Accepted Answer

### The Geometry of Symmetry: Unlocking the Quadrilateral

Welcome, future engineer. Today, we are not just solving a coordinate geometry problem; we are embarking on a journey to uncover the hidden symmetry within two intersecting circles.

Often, when we see equations like $x^2 + y^2 - 2x - 2y - 2 = 0$, our instinct is to dive straight into the algebra. But I want you to pause. Before you touch your pen to paper, look at the structure.

Geometry is the art of seeing the invisible, and today, we will see the square hidden within these circles.

### Phase 1: The Anatomy of the Circles

First, let us decode the identity of our circles. We have two equations:

$$S_1: x^2 + y^2 - 2x - 2y - 2 = 0$$

$$S_2: x^2 + y^2 - 6x - 6y + 14 = 0$$

To understand their behavior, we must find their centers and radii. By completing the square, we transform these into the standard form $(x-h)^2 + (y-k)^2 = r^2$.

For $S_1$, we get $(x-1)^2 + (y-1)^2 = 4$, which tells us the center $C_1$ is at $(1, 1)$ and the radius $r_1$ is $\sqrt{4} = 2$.

For $S_2$, we get $(x-3)^2 + (y-3)^2 = 4$, revealing the center $C_2$ at $(3, 3)$ and the radius $r_2$ is also $\sqrt{4} = 2$.

Notice the elegance here: both circles have the exact same radius. This is the first clue that our quadrilateral will possess a beautiful, balanced symmetry.

### Phase 2: The Rhombus Revelation

Now, imagine the points $P$ and $Q$ where these circles intersect. We are asked to find the area of the quadrilateral $PC_1QC_2$.

Let us connect the vertices: $C_1$ to $P$, $P$ to $C_2$, $C_2$ to $Q$, and $Q$ to $C_1$. Look at the sides of this shape.

$C_1P$ and $C_1Q$ are both radii of the first circle, so they are both $2$. Similarly, $C_2P$ and $C_2Q$ are radii of the second circle, also equal to $2$.

Since all four sides of the quadrilateral are equal to $2$, we have confirmed that $PC_1QC_2$ is a **Rhombus**.

### Phase 3: The Pythagorean Twist

But is it just a rhombus? Let us calculate the length of the diagonal $C_1C_2$ using the distance formula:

$$C_1C_2 = \sqrt{(3-1)^2 + (3-1)^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$$

Now, consider the triangle $\Delta PC_1C_2$. Its sides are $2, 2,$ and $2\sqrt{2}$.

Does this look familiar? If we square the two shorter sides, we get $2^2 + 2^2 = 4 + 4 = 8$. And the square of the longest side is $(2\sqrt{2})^2 = 8$.

By the converse of the Pythagorean theorem, the angle at $P$ must be $90^\circ$! A rhombus with a $90^\circ$ angle is, by definition, a square. We have successfully transformed a complex intersection problem into the simple area calculation of a square with side length $2$.

### The Final Celebration

With the realization that our quadrilateral is a square of side length $2$, the final step is trivial. The area of a square is simply the side length squared:

$$	ext{Area} = 2^2 = 4 	ext{ sq. units}$$

I hope you feel the thrill of this discovery. We didn't need to find the coordinates of $P$ and $Q$ or use complex radical axis equations.

By observing the symmetry and applying the Pythagorean theorem, we found the truth hidden in the geometry. Keep this mindset—always look for the geometric shortcut before diving into the algebraic deep end. You are doing great.

Select Answer:

Visualized Solution

Visualizing the Problem

Analyzing Circle $C_{1}$

Center and Radius of $C_{1}$

Center and Radius of $C_{2}$

Identifying Intersection Points

Forming the Quadrilateral

Analyzing the Sides

Distance Between Centers

Computing $C_{1} C_{2}$

Checking for a Square

Angle at $P$

Area of the Square

The Sigma Insight: Standard and General Equation of a Circle

The Geometry of Symmetry

Phase 1

Phase 2

Phase 3

The Final Celebration

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