This problem combines concepts from ellipses and circles, requiring us to find the radius of a circle that is tangent to an ellipse. The key to solving such problems lies in understanding the geometric condition of tangency and translating it into an algebraic condition, specifically using the discriminant of a quadratic equation.

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1. Fundamental Concept: Tangency and the Discriminant

When a circle is "inscribed" in an ellipse with a fixed center, the "largest" such circle will be tangent to the ellipse. This means the circle touches the ellipse at one or more points without crossing its boundary. Algebraically, if we simultaneously solve the equations of the circle and the ellipse, the resulting equation for the intersection points must have real and equal roots. This condition is mathematically expressed by setting the discriminant ($D$) of the quadratic equation to zero ($D=0$). For a quadratic equation in the form $Ax^2 + Bx + C = 0$, the discriminant is given by $D = B^2 - 4AC$. Setting $D=0$ ensures there is exactly one unique solution, which corresponds to a single point of tangency.

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2. Analyze the Given Ellipse and Convert to Standard Form

The equation of the ellipse is given as: $x^2 + 4y^2 = 36$ To understand its properties (center, semi-axes, orientation), we convert it to the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. We do this by dividing the entire equation by 36: $\frac{x^2}{36} + \frac{4y^2}{36} = \frac{36}{36}$ $\frac{x^2}{6^2} + \frac{y^2}{3^2} = 1$ From this standard form, we can deduce:

• The center of the ellipse is $(0,0)$.
• The semi-major axis is $a = 6$ (along the x-axis, since $6^2 > 3^2$).
• The semi-minor axis is $b = 3$ (along the y-axis). This confirms it's a horizontal ellipse.

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3. Formulate the Circle's Equation

We are given that the center of the circle is $(2,0)$ and its radius is $r$. The general equation of a circle with center $(h,k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$. Substituting the given center $(h,k) = (2,0)$, the equation of our circle is: $(x-2)^2 + y^2 = r^2 \quad \text{(Equation 1)}$

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4. Setting Up for Intersection: Substituting Equations

To find the points where the circle and the ellipse intersect (or touch), we need to solve their equations simultaneously. Our goal is to derive a single equation in terms of one variable (e.g., $x$) and the unknown radius $r$.

From the ellipse equation, it's easiest to isolate $y^2$: $x^2 + 4y^2 = 36$ $4y^2 = 36 - x^2$ $y^2 = \frac{36 - x^2}{4} \quad \text{(Equation 2)}$ Now, substitute this expression for $y^2$ from Equation 2 into the circle's Equation 1. This step eliminates $y$ from the system, leaving us with an equation solely in terms of $x$ and $r$: $(x-2)^2 + \left(\frac{36 - x^2}{4}\right) = r^2$

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5. Deriving the Quadratic Equation in $x$

Next, we expand and simplify the combined equation to obtain a standard quadratic equation in $x$ of the form $Ax^2 + Bx + C = 0$. This form is essential for applying the discriminant condition.

First, expand the term $(x-2)^2$: $(x^2 - 4x + 4) + \frac{36 - x^2}{4} = r^2$ To eliminate the fraction and simplify, multiply the entire equation by 4: $4(x^2 - 4x + 4) + (36 - x^2) = 4r^2$ Distribute the 4 and remove parentheses: $4x^2 - 16x + 16 + 36 - x^2 = 4r^2$ Now, combine like terms ($x^2$ terms, $x$ terms, and constant terms) and move all terms to one side to set the equation to zero: $(4x^2 - x^2) - 16x + (16 + 36) - 4r^2 = 0$ This simplifies to the quadratic equation: $3x^2 - 16x + (52 - 4r^2) = 0 \quad \text{(Equation 3)}$ This quadratic equation gives the $x$-coordinates of the intersection points between the circle and the ellipse.

Tip for JEE: Be very careful with algebraic manipulations, especially when expanding binomials and distributing terms. A single sign error or calculation mistake can lead to an incorrect discriminant.

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6. Applying the Tangency Condition (Discriminant $D=0$)

For the circle to be the largest inscribed circle with its center at $(2,0)$, it must be tangent to the ellipse. This means there should be exactly one distinct $x$-coordinate where they touch. In other words, the quadratic equation (Equation 3) must have real and equal roots.

For a general quadratic equation $Ax^2 + Bx + C = 0$, the condition for real and equal roots (i.e., tangency) is that its discriminant $D = B^2 - 4AC$ must be equal to zero.

From our quadratic equation $3x^2 - 16x + (52 - 4r^2) = 0$:

• Coefficient $A = 3$
• Coefficient $B = -16$
• Constant term $C = (52 - 4r^2)$

Now, set the discriminant to zero: $D = B^2 - 4AC = 0$ $(-16)^2 - 4(3)(52 - 4r^2) = 0$ Calculate the terms: $256 - 12(52 - 4r^2) = 0$ Distribute the $-12$: $256 - (12 \times 52) + (12 \times 4r^2) = 0$ $256 - 624 + 48r^2 = 0$ Combine the constant terms: $-368 + 48r^2 = 0$ $48r^2 = 368$ Why $D=0$ is crucial here:

• If $D > 0$, there would be two distinct real roots for $x$, meaning the circle intersects the ellipse at two distinct points. This would imply the circle is "cutting through" the ellipse, not just touching it internally, and thus wouldn't be the largest inscribed circle for a fixed center.
• If $D < 0$, there would be no real roots for $x$, meaning the circle does not intersect the ellipse at all.

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7. Solving for $12r^2$

From the tangency condition, we have the equation: $48r^2 = 368$ The problem asks for the value of $12r^2$. We can directly obtain this by dividing both sides of the equation by 4: $\frac{48r^2}{4} = \frac{368}{4}$ $12r^2 = 92$

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8. Key Takeaways and JEE Tips

• Geometric to Algebraic Translation: The ability to translate geometric conditions (like tangency) into algebraic equations (like $D=0$) is fundamental in coordinate geometry.
• Systematic Elimination: When dealing with intersecting curves, systematically eliminate one variable to obtain a single equation.
• Standard Forms: Always convert conic section equations to their standard forms to easily identify their properties.
• Discriminant Condition: Remember that $D=0$ is the go-to condition for tangency between a line and a conic, or between two conics (leading to a quadratic in one variable).
• Careful Calculation: Double-check all arithmetic, especially when dealing with squares, multiplications, and signs.
• Read the Question Carefully: Note that the question asked for $12r^2$, not $r$. Avoid extra steps if not necessary, like calculating $r$ itself.

The final answer is $\boxed{\text{92}}$.