The Principle of Shortest Distance: A Foundation for Curve-Point Problems

When asked to find the shortest distance between a point and a curve, a fundamental principle in coordinate geometry guides our approach: The shortest distance from an external point to a curve always lies along the normal to the curve that passes through that point.

This principle is intuitive: if you imagine a string stretching from the point to various points on the curve, the shortest string will be perpendicular to the curve at the point of contact.

In this problem, we need to find the shortest distance from the center of a given circle to a given parabola. Our strategy will leverage this principle:

1. Identify the "point": Determine the coordinates of the center of the circle.
2. Characterize the "curve": Understand the properties of the parabola, specifically its standard form parameter 'a' and the general equation of its normal.
3. Find the specific normal: Determine which normal to the parabola passes through the circle's center. This will give us the slope of that normal.
4. Locate the closest point: Using the slope of the normal, find the exact point on the parabola where this normal intersects it. This is the point on the parabola closest to the circle's center.
5. Calculate the distance: Compute the distance between the circle's center and this closest point on the parabola.
6. Final calculation: Square the distance to fulfill the problem's requirement.

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Step 1: Identifying the Center of the Circle (Our Point of Interest)

The equation of the circle is given in its general form: $x^2 + y^2 - 4x - 16y + 64 = 0$ To find its center, we need to convert this into the standard form of a circle's equation, which is $(x-h)^2 + (y-k)^2 = r^2$, where $(h,k)$ is the center and $r$ is the radius. We achieve this by a technique called completing the square.

First, group the $x$-terms and $y$-terms together: $(x^2 - 4x) + (y^2 - 16y) + 64 = 0$

Now, complete the square for the $x$-terms: To make $x^2 - 4x$ a perfect square trinomial, we add $(\frac{-4}{2})^2 = (-2)^2 = 4$. $(x^2 - 4x + 4) - 4 + (y^2 - 16y) + 64 = 0$ This gives us $(x-2)^2$.

Next, complete the square for the $y$-terms: To make $y^2 - 16y$ a perfect square trinomial, we add $(\frac{-16}{2})^2 = (-8)^2 = 64$. $(x-2)^2 - 4 + (y^2 - 16y + 64) - 64 + 64 = 0$ This gives us $(y-8)^2$.

Substitute these back into the equation: $(x-2)^2 - 4 + (y-8)^2 - 64 + 64 = 0$ Simplify the constant terms: $(x-2)^2 + (y-8)^2 - 4 = 0$ Move the constant to the right side: $(x-2)^2 + (y-8)^2 = 4$

By comparing this with the standard form $(x-h)^2 + (y-k)^2 = r^2$, we can identify the center of the circle, let's call it $C$, as $(h,k) = (2,8)$. The radius of the circle is $r=\sqrt{4}=2$. So, our point of interest is $C(2,8)$.

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Step 2: Characterizing the Parabola and its Normal Equation

The equation of the parabola is given as $y^2 = 4x$. This is in the standard form of a rightward-opening parabola, $y^2 = 4ax$. By comparing $y^2 = 4x$ with $y^2 = 4ax$, we can see that $4a = 4$, which implies $a=1$. This parameter 'a' is crucial for all properties of the parabola.

For a parabola of the form $y^2 = 4ax$, there are several forms for the equation of a normal. When we need to find a normal that passes through a specific external point (like our circle's center $C(2,8)$), the slope form of the normal is generally the most efficient. The equation of a normal to the parabola $y^2 = 4ax$ with slope $m$ is given by: $y = mx - 2am - am^3$ This formula allows us to directly relate the slope of the normal to the coordinates of any point it passes through.

Substitute the value $a=1$ into the normal equation for our specific parabola $y^2=4x$: $y = mx - 2(1)m - (1)m^3$ $y = mx - 2m - m^3$ This is the general equation of a normal to the parabola $y^2=4x$.

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Step 3: Determining the Specific Normal Passing Through the Center

Our objective is to find the normal that passes through the circle's center $C(2,8)$. According to our initial principle, this is the normal that defines the shortest distance. To find the slope $m$ of this specific normal, we substitute the coordinates of $C(2,8)$ into the general normal equation derived in Step 2: $8 = m(2) - 2m - m^3$ $8 = 2m - 2m - m^3$ Notice that the $2m$ terms cancel out, simplifying the equation significantly: $8 = -m^3$ $m^3 = -8$

To solve for $m$, we take the cube root of both sides: $m = \sqrt[3]{-8}$ $m = -2$ This is the unique real slope of the normal to the parabola $y^2=4x$ that passes through the point $(2,8)$. In general, a cubic equation can have up to three real roots, but for $m^3 = \text{constant}$, there is always exactly one real root.

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Step 4: Finding the Point on the Parabola Closest to the Center

Now that we have the slope $m=-2$ of the normal, we need to find the point on the parabola $y^2=4x$ from which this normal originates. This point, let's call it $P$, is the closest point on the parabola to the center $C(2,8)$. For a parabola $y^2 = 4ax$, the coordinates of the foot of the normal (the point on the parabola where the normal with slope $m$ touches it) are given by the parametric form $(am^2, -2am)$.

Using $a=1$ and the calculated slope $m=-2$: The x-coordinate of point $P$ is $x_P = a m^2 = 1 \cdot (-2)^2 = 1 \cdot 4 = 4$. The y-coordinate of point $P$ is $y_P = -2 a m = -2 \cdot 1 \cdot (-2) = 4$.

So, the point $P$ on the parabola closest to the center of the circle is $(4,4)$. (Self-check: Does $P(4,4)$ lie on the parabola $y^2=4x$? Substitute the coordinates: $4^2 = 16$ and $4(4) = 16$. Since $16=16$, the point $P(4,4)$ indeed lies on the parabola.)

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Step 5: Calculating the Shortest Distance 'd'

The shortest distance $d$ is the distance between the point $P(4,4)$ on the parabola and the center of the circle $C(2,8)$. We use the standard distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ Substitute the coordinates of $P(4,4)$ and $C(2,8)$: $d = \sqrt{(4-2)^2 + (4-8)^2}$ $d = \sqrt{(2)^2 + (-4)^2}$ $d = \sqrt{4 + 16}$ $d = \sqrt{20}$

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Step 6: Computing $d^2$

The question specifically asks for the value of $d^2$, not $d$. $d^2 = (\sqrt{20})^2$ $d^2 = 20$

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Key Insights and Common Pitfalls

• Understanding 'a': Always correctly identify the parameter 'a' from the standard form of the parabola ($y^2=4ax$ or $x^2=4ay$). This is the foundation for all subsequent formulas related to tangents, normals, and focal properties.
• Choosing the Right Normal Form: While other forms (like point form or parametric form) exist for normals, the slope form ($y = mx - 2am - am^3$ for $y^2=4ax$) is typically the most direct approach when the normal must pass through a given external point.
• Distance to Center vs. Distance to Circle: This problem explicitly asks for the distance to the center of the circle. If it had asked for the shortest distance to the circle itself, you would calculate the distance to the center (which is $d=\sqrt{20}$) and then subtract the circle's radius ($r=2$) to get $\sqrt{20}-2$. Always read the question carefully!
• Algebraic Precision: Errors in completing the square or solving the cubic equation for 'm' are common. Double-check your calculations. For $m^3 = \text{constant}$, remember there's only one real solution.
• Verifying the Point: After finding the closest point on the parabola, it's a good practice to quickly substitute its coordinates back into the parabola's equation to ensure it actually lies on the curve.

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Summary and Key Takeaway

This problem is an excellent test of fundamental concepts in coordinate geometry, specifically involving circles and parabolas. The core idea is that the shortest distance from a point to a curve is along the normal. The solution systematically applied techniques for finding the center of a circle, recalling the equation of a normal to a parabola, solving for the specific normal's slope, and then using that to find the exact point on the parabola that is closest to the circle's center. Finally, the distance formula was used to calculate the required shortest distance.

The final answer is $\boxed{\text{20}}$.