A square is inscribed in the circle x² + y² - 10x - 6y + 30 = 0. One side of this square is parallel

Q: Wait, why did the problem give us the line $y = x + 3$? I spent ten minutes trying to calculate the angle of rotation!

Ah, the classic JEE distractor! In competitive exams, examiners love to include 'noise' to test your confidence. Because a square is highly symmetric, its centroid will always be the center of the circle regardless of its rotation. The line $y = x + 3$ is mathematically irrelevant to the final sum. Trust your geometric intuition over the urge to use every piece of data provided.

Q: How can we be sure the centroid of the square is the center of the circle?

Think about the symmetry. A square is a cyclic quadrilateral where the diagonals are diameters of the circle. The intersection of the diagonals is the centroid of the square, and since they are diameters, they must intersect at the center of the circle. It is a geometric certainty!

Q: Is it possible to solve this by finding the actual coordinates of the vertices?

Technically, yes, but it is a trap of inefficiency. You would have to find the intersection of the line $y = x + 3 + c$ with the circle, solve for $c$, find the vertices, and then sum their squares. It is algebraically brutal and prone to calculation errors. The summation method we used is elegant, fast, and much less likely to fail.

Question

A square is inscribed in the circle $x^2 + y^2 - 10x - 6y + 30 = 0$. One side of this square is parallel to $y = x + 3$. If $(x_i, y_i)$ are the vertices of the square, then $\sum (x_i^2 + y_i^2)$ is equal to:

Options:

(A) 148

(B) 152

(C) 160

(D) 156

Sigma Percentile · Accepted Answer

### The Art of Seeing Through the Noise

Welcome, fellow traveler on the JEE journey. Today, we are tackling a problem that is less about brute-force calculation and more about the philosophy of geometry.

When you first look at this problem, you see a circle, a square, and a line. Your brain immediately wants to start calculating slopes, finding intersection points, and perhaps even attempting to rotate the coordinate system.

But stop. Take a deep breath. In the world of JEE Advanced, the most elegant solutions often come from realizing what you *don't* need to do.

### Phase 1: Decoding the Circle

We begin with the equation $x^2 + y^2 - 10x - 6y + 30 = 0$. This is the DNA of our circle.

To understand where this circle lives, we must find its center. We compare this to the general form $x^2 + y^2 + 2gx + 2fy + c = 0$.

By matching coefficients, we find $2g = -10$, which gives $g = -5$, and $2f = -6$, which gives $f = -3$. The center $(h, k)$ is $(-g, -f)$, which lands us squarely at $(5, 3)$.

This point is the heartbeat of our problem. Everything else is relative to this coordinate.

### Phase 2: The Great Distractor

The problem mentions that one side of the square is parallel to $y = x + 3$. This is a classic psychological trap.

It is designed to make you panic, to make you think you need to find the specific orientation of the square. But ask yourself: does the sum of the squares of the vertices depend on the rotation of the square?

Imagine the square spinning inside the circle like a wheel. The center of the square remains fixed at the center of the circle. Because the square is perfectly symmetric, its centroid is invariant under rotation. Therefore, the orientation is irrelevant, and we can ignore the line entirely.

### Phase 3: The Algebraic Insight

Now, let us look at the vertices $(x_i, y_i)$. Since they lie on the circle, they must satisfy the circle's equation.

So, for any vertex $i$, we have $x_i^2 + y_i^2 - 10x_i - 6y_i + 30 = 0$. We want the sum $\sum (x_i^2 + y_i^2)$.

Let us rearrange our equation to isolate the terms we need:
$$x_i^2 + y_i^2 = 10x_i + 6y_i - 30$$

This is the key. We are not looking for individual coordinates; we are looking for the sum of these expressions across all four vertices.

### Phase 4: The Power of Summation

We apply the summation operator $\sum_{i=1}^{4}$ to both sides of our equation. This gives us:
$$\sum_{i=1}^{4} (x_i^2 + y_i^2) = 10 \sum_{i=1}^{4} x_i + 6 \sum_{i=1}^{4} y_i - \sum_{i=1}^{4} 30$$

Here, we must be careful. The constant term $30$ is summed four times, so it becomes $120$.

Now, we need the values of $\sum x_i$ and $\sum y_i$. This is where the centroid property shines. The centroid of the square is the center of the circle $(5, 3)$.

By definition, the average of the x-coordinates is the x-coordinate of the center:
$$\frac{\sum_{i=1}^{4} x_i}{4} = 5 \Rightarrow \sum_{i=1}^{4} x_i = 20$$

Similarly, for the y-coordinates:
$$\frac{\sum_{i=1}^{4} y_i}{4} = 3 \Rightarrow \sum_{i=1}^{4} y_i = 12$$

### The Final Victory

We have everything we need. Substituting these sums back into our equation:
$$\sum_{i=1}^{4} (x_i^2 + y_i^2) = 10(20) + 6(12) - 120$$

Calculating this, we find:
$$200 + 72 - 120 = 272 - 120 = 152$$

The result is **152**.

Notice how we never had to solve for a single vertex coordinate? We used the symmetry of the square and the properties of the circle to bypass the tedious algebra. This is the JEE mindset: identify the symmetry, trust the properties, and let the math flow. You have mastered the geometry!

Animated Solution for Mathematics - Circles: A square is inscribed in the circle $x^{2} + y^{2} - 10 x - 6 y + 30 = 0$ . One side of this square is parallel to $y = x + 3$ . If $(x_{i}, y_{i})$ are the vertices of the square, then $\sum (x_{i}^{2} + y_{i}^{2})$ is equal to:

Select Answer:

Visualized Solution

Problem Setup

Circle Equation Analysis

Finding the Center

Vertices on the Circle

Rearranging for $x_{i}^{2} + y_{i}^{2}$

Summing Over All Vertices

Expanding the Summation

The Centroid Concept (JEE Trap)

Calculating $\sum x_{i}$

Calculating $\sum y_{i}$

Substituting the Sums

Final Evaluation

The Sigma Insight: Standard and General Equation of a Circle

The Art of Seeing Through the Noise

Phase 1

Phase 2

Phase 3

Phase 4

The Final Victory

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