This problem requires a strong understanding of both parabolas and circles, and elegantly combines concepts of distance minimization with geometric properties.

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1. Fundamental Principle: Minimum Distance from a Point to a Curve

The core concept for finding the point on a curve that is at the minimum distance from a fixed external point is a geometric one: The line segment connecting the fixed external point to the point of minimum distance on the curve must be perpendicular to the tangent of the curve at that point. In other words, this line segment must be a normal to the curve.

Why is this true? Imagine drawing a series of concentric circles centered at the fixed external point. As the radius of these circles increases, the first circle that just touches (is tangent to) the given curve will identify the point of minimum distance. At the point of tangency, the radius of this circle (which is the line segment connecting the fixed point to the curve) is always perpendicular to the tangent line of the circle. Since the curve is tangent to this circle at that point, the radius must also be perpendicular to the tangent of the curve.

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2. Analyzing the Given Information and Setting Up the Problem

Let's break down the elements provided in the question:

• Parabola: The equation is ${{y^2} = 8x}$.

  • This is a standard parabola of the form ${{y^2} = 4ax}$ opening towards the positive x-axis.
  • Comparing ${{y^2} = 8x}$ with ${{y^2} = 4ax}$, we find ${4a = 8}$, which means ${a = 2}$.
  • Parametric Representation of Point P: For a parabola ${{y^2} = 4ax}$, any general point ${P}$ on it can be conveniently represented using a parameter ${t}$ as $(at^2, 2at)$. This form simplifies calculations involving derivatives and normal lines.
  • Substituting ${a=2}$, the coordinates of point ${P}$ on our parabola are $(2t^2, 2(2)t) = (2t^2, 4t)$.

• Circle C: The equation of the circle is ${x^2} + {\left( {y + 6} \right)^2} = 1$.

  • This is in the standard form ${(x-h)^2} + {(y-k)^2} = r^2$.
  • The center of this circle is ${C(0, -6)}$. The radius is ${r=1}$. Note that the radius of this circle is irrelevant for finding point ${P}$; we only need its center ${C}$.

• Objectives:

  1. Find the coordinates of point ${P}$ on the parabola ${{y^2} = 8x}$ that is at the minimum distance from the center ${C(0, -6)}$.
  2. Once ${P}$ is found, determine the equation of a new circle that passes through ${C(0, -6)}$ and has its center at ${P}$.

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3. Determining Point P on the Parabola

According to our fundamental principle, the line segment ${PC}$ must be normal to the parabola at point ${P(2t^2, 4t)}$. To apply this, we first need to find the equation of the normal to the parabola at ${P}$.

Step 3.1: Find the slope of the tangent at P.

We differentiate the parabola's equation ${{y^2} = 8x}$ with respect to ${x}$ to find the slope of the tangent ${ \frac{dy}{dx} }$: $2y \frac{dy}{dx} = 8$ $\frac{dy}{dx} = \frac{8}{2y} = \frac{4}{y}$ Now, substitute the y-coordinate of point ${P(2t^2, 4t)}$, which is ${y_P = 4t}$, into the derivative to find the slope of the tangent ${m_T}$ at ${P}$: $m_T = \frac{4}{4t} = \frac{1}{t}$ Tip: This formula for ${m_T}$ is valid for ${t \ne 0}$. If ${t=0}$, then ${P}$ is $(0,0)$. The tangent at the origin for ${y^2=8x}$ is the y-axis (a vertical line with an undefined slope). The normal would then be the x-axis (slope 0). However, since ${C(0,-6)}$ is on the y-axis, the normal from ${P(0,0)}$ (the x-axis) would not pass through ${C}$. Thus, ${t \ne 0}$ is expected.

Step 3.2: Find the slope of the normal at P.

The slope of the normal ${m_N}$ is the negative reciprocal of the slope of the tangent: $m_N = -\frac{1}{m_T} = -\frac{1}{1/t} = -t$

Step 3.3: Write the equation of the normal at P.

Using the point-slope form ${(Y - y_1) = m_N(X - x_1)}$ with point ${P(2t^2, 4t)}$ and slope ${m_N = -t}$: $Y - 4t = -t(X - 2t^2)$ $Y - 4t = -tX + 2t^3$ $Y = -tX + 2t^3 + 4t$ This is the equation of the normal to the parabola at point ${P(2t^2, 4t)}$.

Step 3.4: Apply the Minimum Distance Condition to find 't'.

For the distance ${PC}$ to be minimum, the center ${C(0, -6)}$ must lie on the normal we just derived. Substitute the coordinates of ${C(0, -6)}$ into the equation of the normal: $-6 = -t(0) + 2t^3 + 4t$ $-6 = 2t^3 + 4t$ Rearrange this into a standard cubic equation: $2t^3 + 4t + 6 = 0$ Divide the entire equation by 2 to simplify: $t^3 + 2t + 3 = 0$ To find the roots of this cubic equation, we can test integer factors of the constant term (which is 3): ${ \pm 1, \pm 3}$.

• Test ${t = 1}$: ${(1)^3 + 2(1) + 3 = 1 + 2 + 3 = 6 \ne 0}$
• Test ${t = -1}$: ${(-1)^3 + 2(-1) + 3 = -1 - 2 + 3 = 0}$ Since ${t = -1}$ satisfies the equation, it is a root. This means ${(t+1)}$ is a factor of the polynomial ${t^3 + 2t + 3}$. We can perform polynomial division or synthetic division to find the other factor: $(t^3 + 2t + 3) \div (t+1) = t^2 - t + 3$ So, the cubic equation can be factored as: $(t+1)(t^2 - t + 3) = 0$ Now, we need to examine the quadratic factor ${t^2 - t + 3 = 0}$. Let's calculate its discriminant, ${ \Delta = b^2 - 4ac}$: $\Delta = (-1)^2 - 4(1)(3) = 1 - 12 = -11$ Since ${ \Delta < 0}$, the quadratic equation ${t^2 - t + 3 = 0}$ has no real roots. Therefore, the only real value for ${t}$ that satisfies the condition is ${t = -1}$.

Step 3.5: Find the Coordinates of Point P.

Substitute ${t = -1}$ back into the parametric coordinates of ${P(2t^2, 4t)}$: $P = (2(-1)^2, 4(-1))$ $P = (2(1), -4)$ $P = (2, -4)$ So, the point on the parabola at the minimum distance from ${C(0, -6)}$ is ${P(2, -4)}$.

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4. Finding the Equation of the New Circle

We need to find the equation of a circle that:

• Has its center at ${P(2, -4)}$.
• Passes through ${C(0, -6)}$.

Step 4.1: Determine the radius of the new circle.

The radius ${R}$ of this new circle will be the distance between its center ${P(2, -4)}$ and the point it passes through ${C(0, -6)}$. Using the distance formula: ${R = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}}$ $R = \sqrt{(0 - 2)^2 + (-6 - (-4))^2}$ $R = \sqrt{(-2)^2 + (-6 + 4)^2}$ $R = \sqrt{4 + (-2)^2}$ $R = \sqrt{4 + 4}$ $R = \sqrt{8}$ Therefore, the square of the radius is ${R^2 = 8}$.

Step 4.2: Write the equation of the new circle.

The standard equation of a circle with center ${(h, k)}$ and radius ${R}$ is ${(x-h)^2 + (y-k)^2 = R^2}$. Substitute the center ${(h, k) = (2, -4)}$ and ${R^2 = 8}$: $(x - 2)^2 + (y - (-4))^2 = 8$ $(x - 2)^2 + (y + 4)^2 = 8$ Now, expand the terms to get the general form of the circle's equation: $(x^2 - 4x + 4) + (y^2 + 8y + 16) = 8$ $x^2 + y^2 - 4x + 8y + 4 + 16 - 8 = 0$ $x^2 + y^2 - 4x + 8y + 12 = 0$

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5. Final Check and Conclusion

Comparing our derived equation ${{x^2} + {y^2} - 4x + 8y + 12 = 0}$ with the given options: (A) ${{x^2} + {y^2} - {x \over 4} + 2y - 24 = 0}$ (B) ${{x^2} + {y^2} - 4x + 9y + 18 = 0}$ (C) ${{x^2} + {y^2} - 4x + 8y + 12 = 0}$ (D) ${{x^2} + {y^2} - x + 4y - 12 = 0}$

Our result matches option (C).

Key Takeaway: This problem is a classic example of how geometric properties can simplify complex minimization problems in coordinate geometry. The key insight is recognizing that the shortest distance from a point to a curve is along the normal to the curve. This principle is widely applicable. Always remember to leverage parametric forms for curves like parabolas when dealing with tangents, normals, and distances, as they often simplify differentiation and algebraic manipulation. Solving cubic equations by testing integer roots and then factoring is a common technique in such problems.

The final answer is $\boxed{C}$.