The locus of a point, which moves such that the sum of squares of its distances from the points (0,

Q: Wait, why do we use the distance formula squared? Can't we just use the distance formula itself?

Great question! If we used the distance formula without squaring, we would have a sum of square roots: $\sqrt{x^2+y^2} + \sqrt{(x-1)^2+y^2} + \dots = 18$. This would be an algebraic nightmare to solve! By working with the squares of the distances, we eliminate those radicals, turning a complex geometry problem into a simple quadratic equation in $x$ and $y$.

Q: Is there a shortcut? Do I really have to expand all those brackets?

While you could use the property that the sum of squared distances from a point to a set of points is minimized at their centroid, expanding the brackets is actually the safest and most robust way to ensure you don't miss any constant terms. In JEE, accuracy is speed. Trust the algebra!

Q: Why did the equation simplify so nicely? Is that a coincidence?

It is not a coincidence at all! The points $(0,0), (1,0), (0,1),$ and $(1,1)$ are the vertices of a unit square. This symmetry ensures that the linear terms $-2x$ and $-2y$ appear consistently, allowing the equation to form a perfect circle. Symmetry is the hidden language of physics and math.

Q: I got $r=2$, but the answer is $16$. Did I mess up the diameter?

You are actually on the right track! If the radius $r$ is $2$, then the diameter $d$ is $2 \times r = 4$. The question asks for $d^2$, which is $4^2 = 16$. Always re-read the final line of the question—it's the most common place to lose marks in a high-pressure exam!

Q: What if the sum of squares was a different number, like 10? Would it still be a circle?

Yes! As long as the sum of squares is large enough to allow for a real radius, the locus will always be a circle. If the constant were too small, the radius could become imaginary, meaning no such point $P$ exists. You've just discovered the general principle of the 'Circle of Apollonius' family!

Question

The locus of a point, which moves such that the sum of squares of its distances from the points $(0, 0), (1, 0), (0, 1), (1, 1)$ is 18 units, is a circle of diameter $d$. Then $d^2$ is equal to .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a vast, empty coordinate plane. At four specific locations—the origin $(0,0)$, and the corners of a unit square at $(1,0)$, $(0,1)$, and $(1,1)$—you have placed four heavy, immovable anchors.

Now, imagine a point $P(x, y)$ that is tethered to these anchors by invisible, elastic strings. The point $P$ is moving, but it is constrained by a very specific rule: the sum of the squares of its distances to these four anchors must always remain exactly $18$.

This sounds like a chaotic motion, doesn't it? But as we peel back the layers of algebra, we will find that this point is actually tracing a perfect, elegant circle.

### Translating Geometry into Algebra

To capture the motion of $P(x, y)$, we must first translate the physical constraint into the language of mathematics. The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\sqrt{(x-x_1)^2 + (y-y_1)^2}$.

Since we are dealing with the **sum of squares** of these distances, we can discard the square roots. Our constraint equation is:

$$PA^2 + PB^2 + PC^2 + PD^2 = 18$$

Substituting the coordinates of our anchors, we get:

$$[x^2 + y^2] + [(x-1)^2 + y^2] + [x^2 + (y-1)^2] + [(x-1)^2 + (y-1)^2] = 18$$

### The Beauty of Expansion

Looking at four sets of brackets can feel overwhelming, but let’s expand them one by one. We know that $(x-1)^2 = x^2 - 2x + 1$ and $(y-1)^2 = y^2 - 2y + 1$.

When we distribute these into our equation, the terms begin to group themselves. We have four $x^2$ terms and four $y^2$ terms, linear terms $-2x$ and $-2y$ appearing twice each, and a collection of constants.

After expanding and grouping, the equation transforms into:

$$4x^2 + 4y^2 - 4x - 4y + 4 = 18$$

### Finding the Circle's Soul

Now, we simplify. Subtracting $4$ from both sides gives us $4x^2 + 4y^2 - 4x - 4y = 14$. Dividing the entire equation by $4$ reveals the standard form of our circle:

$$x^2 + y^2 - x - y - \frac{7}{2} = 0$$

By comparing this to the general circle equation $x^2 + y^2 + 2gx + 2fy + c = 0$, we identify $g = -\frac{1}{2}$, $f = -\frac{1}{2}$, and $c = -\frac{7}{2}$.

The radius $r$ of any circle is defined by the relationship $r = \sqrt{g^2 + f^2 - c}$. Plugging in our values:

$$r = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)^2 - \left(-\frac{7}{2}\right)} = \sqrt{\frac{1}{4} + \frac{1}{4} + \frac{14}{4}} = \sqrt{\frac{16}{4}} = 2$$

### The Final Victory

We have found that the radius of this path is $2$. The problem asks for the square of the diameter $d$.

Since the diameter $d = 2r$, we have $d = 2(2) = 4$. Therefore, the final result is:

**$d^2 = 4^2 = 16$**

You have successfully mapped the trajectory of the point through the logic of geometry. This result confirms that the point $P$ traces a circle with a diameter of $4$, proving that even complex constraints often hide a beautiful, simple symmetry.

Animated Solution for Mathematics - Circles: The locus of a point, which moves such that the sum of squares of its distances from the points $(0, 0), (1, 0), (0, 1), (1, 1)$ is 18 units, is a circle of diameter $d$ . Then $d^{2}$ is equal to .

Enter Numerical Value:

Visualized Solution

Visualizing the Fixed Points

The Geometric Constraint

Applying the Distance Formula

Setting Up the Equation

Expanding the Terms

Grouping Like Terms

Simplifying to Standard Form

Identifying Circle Parameters

Calculating the Radius

Finding the Final Answer

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

Translating Geometry into Algebra

The Beauty of Expansion

Finding the Circle's Soul

The Final Victory

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