This problem is a classic example of how concepts from coordinate geometry, specifically lines, ellipses, and circles, are integrated. Our goal is to find the length of the chord formed by the intersection of a given line and an ellipse, which then serves as the diameter of a circle. Finally, we need to calculate a specific expression involving the radius of this circle.

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1. Core Concepts and Formulas

Before we dive into the solution, let's lay down the fundamental mathematical tools we'll be using. Understanding these concepts is crucial for solving problems of this type.

• Equation of a Line: A straight line can be represented as $y = mx + c$, where $m$ is the slope and $c$ is the y-intercept. In our problem, the line is $y = x + 1$, so $m=1$ and $c=1$.
• Equation of an Ellipse: A standard ellipse centered at the origin is given by $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Our given ellipse is $\frac{x^2}{4} + \frac{y^2}{2} = 1$, where $a^2=4$ and $b^2=2$.
• Intersection of a Line and an Ellipse: When a line intersects an ellipse, the points of intersection satisfy both the line's equation and the ellipse's equation. To find these points, we substitute the expression for $y$ (or $x$) from the line equation into the ellipse equation. This eliminates one variable and results in a quadratic equation in the other variable (either $x$ or $y$). The roots of this quadratic equation will be the coordinates of the intersection points.
• Quadratic Equation and Vieta's Formulas: For a quadratic equation of the form $Ax^2 + Bx + C = 0$, if its roots are $\alpha$ and $\beta$, then:
  • Sum of roots: $\alpha + \beta = -\frac{B}{A}$
  • Product of roots: $\alpha\beta = \frac{C}{A}$
  • Important Identity: The square of the difference of the roots can be expressed as: $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$. This identity is incredibly useful because it allows us to find the squared difference of roots without explicitly calculating the roots themselves, which often involve square roots and complex arithmetic, thus simplifying calculations significantly.
• Distance Formula: The distance between two points $P(x_1, y_1)$ and $Q(x_2, y_2)$ in a Cartesian plane is given by $PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. Often, it's more convenient and algebraically simpler to work with the squared distance: $PQ^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$.
• Circle Properties: If a line segment PQ is the diameter of a circle, then its length is twice the radius $r$. So, $PQ = 2r$, which implies $PQ^2 = (2r)^2 = 4r^2$. This relationship allows us to connect the length of the chord to the circle's radius.

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2. Step-by-Step Solution

Let's break down the problem into manageable steps, explaining the rationale behind each action.

Step 1: Formulate the Quadratic Equation for the x-coordinates of the Intersection Points

Goal: Find the x-coordinates of the points P and Q where the line intersects the ellipse. Method: Substitute the equation of the line into the equation of the ellipse. This eliminates $y$ and leaves an equation solely in terms of $x$, whose solutions will be the x-coordinates of P and Q.

Given line: $y = x + 1$ Given ellipse: $\frac{x^2}{4} + \frac{y^2}{2} = 1$

Substitute $y = x + 1$ into the ellipse equation: $\frac{x^2}{4} + \frac{(x + 1)^2}{2} = 1$

Now, we need to simplify this equation into the standard quadratic form $Ax^2 + Bx + C = 0$. To clear the denominators and work with integers, we multiply the entire equation by the least common multiple (LCM) of 4 and 2, which is 4: $4 \left( \frac{x^2}{4} \right) + 4 \left( \frac{(x + 1)^2}{2} \right) = 4 \times 1$ $x^2 + 2(x + 1)^2 = 4$ Next, we expand the binomial term $(x + 1)^2$ using the identity $(a+b)^2 = a^2+2ab+b^2$: $x^2 + 2(x^2 + 2x + 1) = 4$ Distribute the 2 into the parenthesis: $x^2 + 2x^2 + 4x + 2 = 4$ Combine like terms ($x^2$ terms) and move all terms to one side to form the quadratic equation: $(1+2)x^2 + 4x + 2 - 4 = 0$ $3x^2 + 4x - 2 = 0$ This is our quadratic equation. Let its roots be $\alpha$ and $\beta$. These roots are precisely the x-coordinates of the intersection points P and Q.

Step 2: Express the Coordinates of P and Q

Goal: Define the coordinates of points P and Q using the roots found in Step 1. Method: Since the x-coordinates are $\alpha$ and $\beta$, we can find their corresponding y-coordinates by substituting $\alpha$ and $\beta$ into the line equation $y = x + 1$.

So, the coordinates of the two points are: $P(\alpha, \alpha + 1)$ $Q(\beta, \beta + 1)$

Step 3: Calculate the Length of the Chord PQ (which is the Diameter of the Circle)

Goal: Find the length of the segment PQ, which is the diameter of the circle, and relate it to the radius $r$. Method: Use the distance formula for $PQ^2$ to avoid square roots initially, and then use the relationship $PQ = 2r$.

The distance formula for $PQ$ is $PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. Substituting the coordinates of P and Q: $PQ = \sqrt{(\beta - \alpha)^2 + ((\beta + 1) - (\alpha + 1))^2}$ Let's simplify the difference in y-coordinates: $(\beta + 1) - (\alpha + 1) = \beta + 1 - \alpha - 1 = \beta - \alpha$ Notice that because the line has a slope of 1 ($y = x + 1$), the difference in y-coordinates is exactly equal to the difference in x-coordinates. This is a common simplification for lines with $m=1$ or $m=-1$.

Substitute this back into the distance formula: $PQ = \sqrt{(\beta - \alpha)^2 + (\beta - \alpha)^2}$ $PQ = \sqrt{2(\beta - \alpha)^2}$ Since $PQ$ is the diameter of the circle, its length is $2r$. Therefore, $PQ^2 = (2r)^2 = 4r^2$. Let's square the expression for $PQ$: $PQ^2 = \left(\sqrt{2(\beta - \alpha)^2}\right)^2$ $PQ^2 = 2(\beta - \alpha)^2$ Equating this with $4r^2$: $4r^2 = 2(\beta - \alpha)^2$ Dividing both sides by 2, we get a direct relationship between $r^2$ and the squared difference of the roots: $2r^2 = (\beta - \alpha)^2$ Since $(\beta - \alpha)^2$ is the same as $(\alpha - \beta)^2$, we can write: $2r^2 = (\alpha - \beta)^2$ This is a crucial intermediate result, as it connects the radius to the roots of our quadratic equation.

Step 4: Use Vieta's Formulas to Find $(\alpha - \beta)^2$

Goal: Calculate the value of $(\alpha - \beta)^2$ using the coefficients of the quadratic equation. Method: Apply Vieta's formulas to find the sum and product of the roots of $3x^2 + 4x - 2 = 0$, then use the identity $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$.

From the quadratic equation $3x^2 + 4x - 2 = 0$ (obtained in Step 1), we identify the coefficients: $A = 3$, $B = 4$, $C = -2$.

Using Vieta's formulas:

• Sum of roots: $\alpha + \beta = -\frac{B}{A} = -\frac{4}{3}$
• Product of roots: $\alpha\beta = \frac{C}{A} = \frac{-2}{3}$

Now, we use the identity $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$: $(\alpha - \beta)^2 = \left(-\frac{4}{3}\right)^2 - 4\left(-\frac{2}{3}\right)$ First, calculate the square and the product: $(\alpha - \beta)^2 = \frac{(-4)^2}{3^2} - \left(-\frac{8}{3}\right)$ $(\alpha - \beta)^2 = \frac{16}{9} + \frac{8}{3}$ To add these fractions, we find a common denominator, which is 9: $(\alpha - \beta)^2 = \frac{16}{9} + \frac{8 \times 3}{3 \times 3}$ $(\alpha - \beta)^2 = \frac{16}{9} + \frac{24}{9}$ $(\alpha - \beta)^2 = \frac{16 + 24}{9} = \frac{40}{9}$

Step 5: Calculate the Required Value $(3r)^2$

Goal: Use the value of $(\alpha - \beta)^2$ to find $r^2$ and then calculate $(3r)^2$. Method: Substitute the value of $(\alpha - \beta)^2$ into the relationship derived in Step 3, then perform the final calculation.

We have the relationship from Step 3: $2r^2 = (\alpha - \beta)^2$ Substitute the value of $(\alpha - \beta)^2 = \frac{40}{9}$ from Step 4: $2r^2 = \frac{40}{9}$ Now, we need to find $r^2$. Divide both sides by 2: $r^2 = \frac{40}{9 \times 2} = \frac{40}{18}$ Simplify the fraction by dividing both numerator and denominator by 2: $r^2 = \frac{20}{9}$ The question asks for the value of $(3r)^2$. We can calculate this by first squaring $3r$: $(3r)^2 = 3^2 \times r^2 = 9r^2$ Now, substitute the value of $r^2$: $(3r)^2 = 9 \times \frac{20}{9}$ The 9 in the numerator and denominator cancel out: $(3r)^2 = 20$

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3. Tips for Success and Common Pitfalls

• Algebraic Accuracy is Paramount: A single sign error or miscalculation with fractions can invalidate the entire solution. Double-check each step, especially when expanding binomials, distributing terms, and combining fractions.
• Don't Rush to Solve the Quadratic: It's a common mistake to immediately try to find the exact roots of the quadratic equation using the quadratic formula. However, this often leads to complex expressions involving square roots, which makes subsequent calculations (like the distance formula) much more tedious and error-prone. The identity $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$ is designed precisely to avoid this, leveraging Vieta's formulas instead. This is a key time-saving technique in competitive exams.
• Distance Squared vs. Distance: Pay attention to whether you need $PQ$ or $PQ^2$. In problems involving radii and diameters, working with squared distances ($PQ^2 = 4r^2$) often simplifies the algebra significantly, avoiding square roots until the very end (if needed).
• Understanding the Line's Slope: Notice how the line $y=x+1$ (slope $m=1$) simplified the distance formula step. For a general line $y=mx+c$, the difference in y-coordinates is $y_2 - y_1 = (mx_2+c) - (