A point P moves so that the sum of squares of its distances from the points (1, 2) and (-2, 1) is 14

Q: Wait, why do we reject the negative root here or why is the distance formula not giving me a square root?

Great observation! The beauty of this problem lies in the condition $PQ_1^2 + PQ_2^2 = 14$. Because we are dealing with the *square* of the distance, the square root from the standard distance formula $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ is squared away, leaving us with a clean algebraic expression. It is a gift from the problem setter to keep our calculations elegant!

Q: How do I know for sure that the locus is a circle?

Look at the final simplified equation: $x^2 + y^2 + x - 3y - 2 = 0$. A general second-degree equation represents a circle if the coefficients of $x^2$ and $y^2$ are equal and there is no $xy$ term. Here, both coefficients are $1$, and there is no $xy$ term, confirming it is indeed a circle.

Q: Why is the area of the quadrilateral just $\frac{1}{2} \times AB \times CD$?

This is a powerful geometric property! For any quadrilateral where the diagonals are perpendicular, the area is exactly half the product of the lengths of the diagonals. Since our diagonals $AB$ and $CD$ lie on the $x$ and $y$ axes, they are inherently perpendicular, allowing us to use this shortcut instead of complex coordinate geometry formulas.

Question

A point $P$ moves so that the sum of squares of its distances from the points $(1, 2)$ and $(-2, 1)$ is $14$. Let $f(x, y) = 0$ be the locus of $P$, which intersects the $x$-axis at the points $A, B$ and the $y$-axis at the point $C, D$. Then the area of the quadrilateral $ACBD$ is equal to

Options:

(A) $\frac{9}{2}$

(B) $\frac{3\sqrt{17}}{2}$

(C) $\frac{3\sqrt{17}}{4}$

(D) $9$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a point $P(x, y)$ moving on a coordinate plane such that the sum of the squares of its distances from two fixed anchors, $Q_1(1, 2)$ and $Q_2(-2, 1)$, is constant. The given condition is expressed as:

$$PQ_1^2 + PQ_2^2 = 14$$

Using the distance formula $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$, we substitute the coordinates of $P$, $Q_1$, and $Q_2$ into the condition:

$$(x - 1)^2 + (y - 2)^2 + (x + 2)^2 + (y - 1)^2 = 14$$

### The Master Equation

To simplify this expression, we expand each squared term individually:

$$(x^2 - 2x + 1) + (y^2 - 4y + 4) + (x^2 + 4x + 4) + (y^2 - 2y + 1) = 14$$

Grouping the like terms together, we obtain:

$$2x^2 + 2y^2 + 2x - 6y + 10 = 14$$

Subtracting $14$ from both sides and dividing the entire equation by $2$, we arrive at the standard form of the circle:

$$x^2 + y^2 + x - 3y - 2 = 0$$

### Identifying the Geometry

The equation $x^2 + y^2 + x - 3y - 2 = 0$ features identical coefficients for $x^2$ and $y^2$ with no $xy$ term, confirming that the path of $P$ is a **circle**. We now determine the intercepts to find the dimensions of the quadrilateral formed by them.

To find the $x$-intercepts ($A$ and $B$), we set $y = 0$:

$$x^2 + x - 2 = 0 \Rightarrow (x + 2)(x - 1) = 0$$

This yields $x = -2$ and $x = 1$. Thus, the length of the horizontal diagonal is $AB = |1 - (-2)| = 3$.

To find the $y$-intercepts ($C$ and $D$), we set $x = 0$:

$$y^2 - 3y - 2 = 0$$

Applying the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we get:

$$y = \frac{3 \pm \sqrt{9 - 4(1)(-2)}}{2} = \frac{3 \pm \sqrt{17}}{2}$$

The length of the vertical diagonal is $CD = \frac{3 + \sqrt{17}}{2} - \frac{3 - \sqrt{17}}{2} = \sqrt{17}$.

### Final Calculation

Since the $x$ and $y$ axes are perpendicular, the diagonals of the quadrilateral are perpendicular. The area of a quadrilateral with perpendicular diagonals is given by:

$$	ext{Area} = \frac{1}{2} 	imes d_1 	imes d_2$$

Substituting our calculated lengths:

$$	ext{Area} = \frac{1}{2} 	imes 3 	imes \sqrt{17} = \frac{3\sqrt{17}}{2}$$

The final area of the quadrilateral is **$\frac{3\sqrt{17}}{2}$**.

Select Answer:

Visualized Solution

Visualizing the Setup

The Given Condition

Applying the Distance Formula

Setting up the Equation

Expanding the Squares

Simplifying to Standard Form

Identifying the Locus

Finding Intersections with X-axis

Solving for Points A and B

Finding Intersections with Y-axis

Solving for Points C and D

Visualizing the Quadrilateral

Calculating the Area

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Master Equation

Identifying the Geometry

Final Calculation

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