This problem requires a systematic approach involving the properties of tangents to parabolas and circles. We need to find a line that is tangent to both given curves, determine the points where it touches each curve, and then calculate the squared distance between these two points.

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1. Understanding the Problem and Initial Strategy

We are given two curves:

1. A circle: $x^2+y^2=8$
2. A parabola: $y^2=16x$

We need to find a common tangent line to these two curves. Let this tangent line touch the parabola at point $Q$ and the circle at point $R$. Our ultimate goal is to calculate the squared distance $(QR)^2$. The information that the common tangents are drawn from a point $A$ on the $x$-axis serves as a consistency check for our tangent equations, but it's not strictly necessary for finding the common tangents themselves.

Our strategy will be:

1. Write down the general equation of a tangent to the parabola in terms of its slope.
2. Apply the tangency condition for this general line to the circle to find the specific slope(s) of the common tangent(s).
3. Choose one of the common tangent lines and find its point of tangency on the parabola (point $Q$).
4. Using the same tangent line, find its point of tangency on the circle (point $R$).
5. Calculate the squared distance between $Q$ and $R$.

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2. Finding the Equation of the Common Tangent

2.1. General Tangent to the Parabola

The given parabola is $y^2 = 16x$. The standard form of a parabola opening to the right is $y^2 = 4ax$. Comparing $y^2 = 16x$ with $y^2 = 4ax$, we can identify the parameter $a$: $4a = 16 \implies a = 4$.

The equation of a tangent to the parabola $y^2 = 4ax$ with slope $m$ is given by the slope-form formula: $y = mx + \frac{a}{m}$ Why this formula? This is a standard result derived by setting the discriminant of the quadratic formed by substituting $y=mx+c$ into $y^2=4ax$ to zero. Substituting $a=4$ into the tangent equation, we get the general equation of a tangent to $y^2=16x$: $y = mx + \frac{4}{m} \quad \text{ (Equation 1)}$ This equation represents any line that is tangent to the parabola $y^2=16x$. Now, we need to find the specific value(s) of $m$ for which this line is also tangent to the circle.

2.2. Tangency Condition for the Circle

The given circle is $x^2+y^2=8$. The standard form of a circle centered at the origin is $x^2+y^2=r^2$. Comparing $x^2+y^2=8$ with $x^2+y^2=r^2$, we find the center is $(0,0)$ and the radius squared is $r^2=8$. So, the radius is $r = \sqrt{8} = 2\sqrt{2}$.

For a line to be tangent to a circle, the perpendicular distance from the center of the circle to the line must be equal to the radius of the circle. First, rewrite Equation 1 in the general linear form $Ax+By+C=0$: $mx - y + \frac{4}{m} = 0$ Here, $A=m$, $B=-1$, and $C=\frac{4}{m}$. The center of the circle is $(x_0, y_0) = (0,0)$. The formula for the perpendicular distance from a point $(x_0, y_0)$ to a line $Ax+By+C=0$ is $\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$. Applying this condition: $\frac{\left|m(0) - 1(0) + \frac{4}{m}\right|}{\sqrt{m^2 + (-1)^2}} = r$ $\frac{\left|\frac{4}{m}\right|}{\sqrt{m^2+1}} = 2\sqrt{2}$ Why the absolute value? Distance is always non-negative. Squaring both sides will remove the absolute value and the square root, making it easier to solve for $m$. $\frac{\left(\frac{4}{m}\right)^2}{m^2+1} = (2\sqrt{2})^2$ $\frac{16}{m^2(m^2+1)} = 8$ Now, we solve for $m$: $16 = 8m^2(m^2+1)$ Divide both sides by 8: $2 = m^2(m^2+1)$ $2 = m^4 + m^2$ Rearrange into a quadratic form by setting the equation to zero: $m^4 + m^2 - 2 = 0$ Let $k=m^2$ to simplify the quadratic equation: $k^2 + k - 2 = 0$ Factor the quadratic equation: $(k+2)(k-1) = 0$ This gives two possible values for $k$: $k=-2$ or $k=1$. Since $k=m^2$, and $m$ represents a real slope, $m^2$ must be non-negative. Therefore, we discard $m^2=-2$ because it would imply $m$ is imaginary. So, we must have $m^2=1$, which implies $m = \pm 1$.

These are the slopes of the common tangents. We can choose either value of $m$ to find the points of tangency. Let's choose $m=1$. Substituting $m=1$ into Equation 1: $y = (1)x + \frac{4}{1} \implies \mathbf{y = x+4}$ This is one of the common tangents. If we chose $m=-1$, we would get $y = -x-4$. Both tangents would yield the same squared distance between their respective points of tangency.

(Optional Check: The problem states common tangents are drawn from a point A on the x-axis. For $y=x+4$, if A is $(x_A, 0)$, then $0 = x_A+4 \implies x_A=-4$. So $A(-4,0)$ is the point from which these tangents originate. This is consistent with the problem statement.)

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3. Finding the Points of Tangency

Now we use the common tangent $y=x+4$ to find the points where it touches the parabola and the circle.

3.1. Point Q on the Parabola ($y^2=16x$)

To find the point of tangency, we substitute the equation of the tangent line ($y=x+4$) into the equation of the parabola ($y^2=16x$). $(x+4)^2 = 16x$ Why substitute? The points of tangency are the intersection points of the line and the curve. Since a tangent touches at exactly one point, the resulting quadratic equation in $x$ (or $y$) will have exactly one solution (a repeated root). Expand the left side: $x^2 + 8x + 16 = 16x$ Rearrange into a standard quadratic equation: $x^2 - 8x + 16 = 0$ This is a perfect square trinomial: $(x-4)^2 = 0$ This implies $x=4$. Now, substitute $x=4$ back into the tangent equation $y=x+4$ to find the $y$-coordinate: $y = 4+4 = 8$ So, the point of tangency on the parabola is $\mathbf{Q(4,8)}$.

3.2. Point R on the Circle ($x^2+y^2=8$)

Similarly, we substitute the equation of the tangent line ($y=x+4$) into the equation of the circle ($x^2+y^2=8$). $x^2 + (x+4)^2 = 8$ Expand the squared term: $x^2 + (x^2 + 8x + 16) = 8$ Combine like terms: $2x^2 + 8x + 16 = 8$ Move the constant term to the left side to set the equation to zero: $2x^2 + 8x + 8 = 0$ Divide the entire equation by 2 to simplify: $x^2 + 4x + 4 = 0$ This is also a perfect square trinomial: $(x+2)^2 = 0$ This implies $x=-2$. Now, substitute $x=-2$ back into the tangent equation $y=x+4$ to find the $y$-coordinate: $y = -2+4 = 2$ So, the point of tangency on the circle is $\mathbf{R(-2,2)}$.

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