This solution will guide you through a comprehensive breakdown of the problem, focusing on the underlying concepts, step-by-step calculations, and crucial insights needed to solve it effectively.

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1. Understanding the Ellipse Equation and Identifying Axes

The first crucial step is to correctly interpret the given ellipse equation and determine the coordinates of its endpoints on the axes. This involves converting the equation into its standard form.

• Key Concept: Standard Form of an Ellipse An ellipse centered at the origin $(0,0)$ has the standard equation: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ Here, $a$ is the length of the semi-major/minor axis along the x-axis, and $b$ is the length of the semi-major/minor axis along the y-axis.

  • If $a > b$, the major axis lies along the x-axis, and the minor axis along the y-axis. The vertices are $(\pm a, 0)$ and $(0, \pm b)$.
  • If $b > a$, the major axis lies along the y-axis, and the minor axis along the x-axis. The vertices are $(\pm a, 0)$ and $(0, \pm b)$.

• Given Ellipse Equation: We are given the equation of the ellipse $\mathrm{E}_{k}$: $k x^{2}+k^{2} y^{2}=1$

• Step-by-Step Conversion to Standard Form: To match the standard form, we need $x^2$ and $y^2$ to have a coefficient of 1. We achieve this by dividing the terms by their respective coefficients: $\frac{x^2}{\frac{1}{k}} + \frac{y^2}{\frac{1}{k^2}} = 1$

• Identifying Semi-Axes: By comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$:

  • $a^2 = \frac{1}{k} \implies a = \frac{1}{\sqrt{k}}$ (This is the semi-axis length along the x-axis).
  • $b^2 = \frac{1}{k^2} \implies b = \frac{1}{k}$ (This is the semi-axis length along the y-axis).
  • Why these values? These values directly correspond to the denominators in the standard form, representing the squares of the semi-axis lengths.

• Determining Major and Minor Axes: We need to compare $a$ and $b$: For $k=1$, $a = \frac{1}{\sqrt{1}} = 1$ and $b = \frac{1}{1} = 1$. In this case, $a=b$, so $\mathrm{E}_1$ is a circle. For $k > 1$ (e.g., $k=2$), we have $\sqrt{k} < k$. This implies $\frac{1}{\sqrt{k}} > \frac{1}{k}$. So, for $k > 1$, $a > b$.

  • The x-axis is the major axis with endpoints $\left(\pm a, 0\right) = \left(\pm \frac{1}{\sqrt{k}}, 0\right)$.
  • The y-axis is the minor axis with endpoints $\left(0, \pm b\right) = \left(0, \pm \frac{1}{k}\right)$.

  The four endpoints of the axes are:

  • $A_1 \left(\frac{1}{\sqrt{k}}, 0\right)$, $A_2 \left(-\frac{1}{\sqrt{k}}, 0\right)$ (on the x-axis/major axis)
  • $B_1 \left(0, \frac{1}{k}\right)$, $B_2 \left(0, -\frac{1}{k}\right)$ (on the y-axis/minor axis)

  Tip: Always verify which axis is major and which is minor. A common mistake is to assume the x-axis is always the major axis. In this case, $k>1$ implies $a>b$, confirming the x-axis as major.

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2. Forming the Chords and Determining the Circle's Center

Next, we need to identify the four chords mentioned in the problem and use their properties to deduce the center of the circle $C_k$.

• Defining the Chords: The problem states that $C_k$ is a circle that touches four chords. Each chord joins an endpoint on the minor axis with an endpoint on the major axis. These are lines connecting points like $(\pm a, 0)$ and $(0, \pm b)$.

• Key Concept: Equation of a Line in Intercept Form The equation of a line that passes through the x-intercept $(X_0, 0)$ and the y-intercept $(0, Y_0)$ is given by: $\frac{x}{X_0} + \frac{y}{Y_0} = 1$ Why this form? We have the exact intercepts for each chord, making this the most direct way to write their equations.

• Step-by-Step Equation of a Chord: Let's consider one specific chord, for instance, the one joining $A_1\left(\frac{1}{\sqrt{k}}, 0\right)$ and $B_1\left(0, \frac{1}{k}\right)$. Here, $X_0 = \frac{1}{\sqrt{k}}$ and $Y_0 = \frac{1}{k}$. Substituting these values into the intercept form: $\frac{x}{\frac{1}{\sqrt{k}}} + \frac{y}{\frac{1}{k}} = 1$ Simplifying this equation: $\sqrt{k}x + ky = 1$ To use the distance formula later, we rewrite this in the general form $Ax+By+C=0$: $\sqrt{k}x + ky - 1 = 0$

• The Four Chords: By considering all combinations of endpoints, the four chords are:

  1. Joining $\left(\frac{1}{\sqrt{k}}, 0\right)$ and $\left(0, \frac{1}{k}\right)$: $\sqrt{k}x + ky - 1 = 0$
  2. Joining $\left(-\frac{1}{\sqrt{k}}, 0\right)$ and $\left(0, \frac{1}{k}\right)$: $-\sqrt{k}x + ky - 1 = 0$
  3. Joining $\left(\frac{1}{\sqrt{k}}, 0\right)$ and $\left(0, -\frac{1}{k}\right)$: $\sqrt{k}x - ky - 1 = 0$
  4. Joining $\left(-\frac{1}{\sqrt{k}}, 0\right)$ and $\left(0, -\frac{1}{k}\right)$: $-\sqrt{k}x - ky - 1 = 0$

• Determining the Center of Circle $C_k$: Observe the equations of these four lines. They form a figure (specifically, a rhombus) that is perfectly symmetric with respect to both the x-axis and the y-axis. Why is this important? If a circle touches all four sides of such a symmetric figure, its center must lie at the intersection of the axes of symmetry. In this case, the intersection of the x-axis and y-axis is the origin $(0,0)$. Therefore, the circle $C_k$ is centered at the origin $(0,0)$.

  Tip: A quick sketch can often reveal symmetry. If a circle is tangent to lines $L_1, L_2, L_3, L_4$ and these lines form a figure symmetric about the x-axis and y-axis, the center of the circle must be $(0,0)$.

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3. Calculating the Radius $r_k$ of Circle $C_k$

Since the circle $C_k$ is centered at the origin $(0,0)$ and touches all four chords, its radius $r_k$ is simply the perpendicular distance from the origin to any one of these chords.

• Key Concept: Perpendicular Distance from a Point to a Line The perpendicular distance $d$ from a point $(x_1, y_1)$ to a line $Ax+By+C=0$ is given by the formula: $d = \frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$ Why this formula? This directly calculates the shortest distance from a point to a line, which is precisely the radius of a circle tangent to that line from a given center.

• Step-by-Step Calculation of $r_k$: Let's use the first chord equation: $\sqrt{k}x + ky - 1 = 0$. Here, $A = \sqrt{k}$, $B = k$, and $C = -1$. The point is the center of the circle, $(x_1, y_1) = (0,0)$. Substituting these values into the distance formula: $r_k = \frac{|\sqrt{k}(0) + k(0) - 1|}{\sqrt{(\sqrt{k})^2 + (k)^2}}$ $r_k = \frac{|-1|}{\sqrt{k + k^2}}$ $r_k = \frac{1}{\sqrt{k^2 + k}}$

• Calculating $1/r_k^2$: The problem asks for the sum of $\frac{1}{r_k^2}$, not $r_k$. So, let's find this expression: $\frac{1}{r_k^2} = \left(\frac{1}{\sqrt{k^2 + k}}\right)^{-2}$ $\frac{1}{r_k^2} = \left(\sqrt{k^2 + k}\right)^2$ $\frac{1}{r_k^2} = k^2 + k$

  Common Mistake: Forgetting to square the numerator and denominator correctly, or calculating $r_k$ and then trying to sum it, instead of finding $1/r_k^2$ first. Always double-check what quantity is required for the summation.

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4. Evaluating the Summation

Finally, we need to calculate the sum of the derived expression for $k$ from 1 to 20.

• Substituting the Expression for $1/r_k^2$: The problem asks for the value of $\sum_{k=1}^{20} \frac{1}{r_{k}^{2}}$. Substituting our result from the previous step: $\sum_{k=1}^{20} \frac{1}{r_{k}^{2}} = \sum_{k=1}^{20} (k^2 + k)$

• Separating the Summation: The sum of a sum is the sum of the individual sums: $\sum_{k=1}^{20} (k^2 + k) = \sum_{k=1}^{20} k^2 + \sum_{k=1}^{20} k$

• Key Concepts: Standard Summation Formulas We will use two well-known formulas for sums of series:

  1. Sum of the first $n$ natural numbers: $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$
  2. Sum of the squares of the first $n$ natural numbers: $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$ Why these formulas? These are standard results that allow for quick calculation of such arithmetic series, avoiding tedious direct summation.

• Step-by-Step Calculation: For our problem, $n=20$.

  1. Calculate $\sum_{k=1}^{20} k$: $\sum_{k=1}^{20} k = \frac{20(20+1)}{2} = \frac{20 \times 21}{2} = 10 \times 21 = 210$

  2. Calculate $\sum_{k=1}^{20} k^2$: $\sum_{k=1}^{20} k^2 = \frac{20(20+1)(2 \times 20+1)}{6} = \frac{20 \times 21 \times 41}{6}$ To simplify, divide 20 by 2 and 21 by 3 (or 6 by 6): $= \frac{(20 \times 21)}{6} \times 41 = \frac{420}{6} \times 41 = 70 \times 41$ $= 2870$

  3. Add the two sums: $\sum_{k=1}^{20} (k^2 + k) = 2870 + 210 = 3080$

• Final Result: The value of $\sum_{k=1}^{20} \frac{1}{r_{k}^{2}}$ is $\boxed{3080}$.

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Summary and Key Takeaways

This problem is a fantastic example of how multiple concepts from coordinate geometry and series are interwoven.

1. Standard Forms are Your Friends: Always convert given equations (like the ellipse) into their standard forms to easily extract key parameters.
2. Leverage Symmetry: Recognizing geometric symmetry (like the rhombus formed by the chords) can drastically