This problem is a comprehensive test of coordinate geometry concepts, integrating knowledge of straight lines, parabolas, and circles. We will systematically break down the problem into smaller, manageable steps to arrive at the solution.

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Overall Strategy

Our approach will be structured as follows:

1. Determine the slope of the tangent line: The tangent line is perpendicular to a given line, which allows us to find its slope using the relationship between perpendicular slopes.
2. Formulate the equation of the tangent line: Using the identified slope and the properties of the given parabola, we will write the equation of the tangent line.
3. Find the point of tangency (P): This point is the intersection of the tangent line and the parabola, which we will find by solving their equations simultaneously.
4. Locate the center of the circle: From the given general equation of the circle, we will extract its center coordinates.
5. Calculate the distance: Finally, we will use the distance formula to find the distance between point P and the center of the circle.

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1. Finding the Slope of the Tangent Line

• Key Concept: If two lines are perpendicular, the product of their slopes is $-1$. That is, if $m_1$ and $m_2$ are the slopes of two perpendicular lines, then $m_1 \cdot m_2 = -1$.

• Step-by-step Working:

  • Given Line: We are provided with the line $2x - y = 10$.
  • Why this step? To find the slope of the tangent line, we first need to know the slope of the line it is perpendicular to.
  • Finding the slope of the given line: To easily identify its slope, we rewrite the equation in the standard slope-intercept form, $y = mx + c$, where $m$ is the slope. $2x - y = 10$ $y = 2x - 10$ By comparing this with $y=mx+c$, we find that the slope of this given line is $m_1 = 2$.
  • Finding the slope of the perpendicular tangent line: Let the slope of the tangent line at point P be $m_T$. Since this tangent line is perpendicular to the given line ($2x - y = 10$), their slopes must satisfy the condition for perpendicular lines: $m_1 \cdot m_T = -1$ Substituting $m_1 = 2$: $2 \cdot m_T = -1$ $m_T = -\frac{1}{2}$
  • Result: The slope of the tangent line at point P is $m = -\frac{1}{2}$.

• Tips for Success: Always be meticulous with signs when calculating the negative reciprocal. A common error is to forget the negative sign or to only take the reciprocal without changing the sign.

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2. Finding the Equation of the Tangent Line to the Parabola

• Key Concept: For a parabola with its axis parallel to the x-axis, given by the equation $y^2 = 4a(x-h)$, the equation of a tangent line with a slope $m$ is a standard formula: $y = m(x-h) + \frac{a}{m}$ This formula is derived from the condition of tangency (discriminant of the quadratic formed by intersection is zero) or using differentiation.

• Step-by-step Working:

  • Given Parabola: The equation of the parabola is $y^2 = 4(x-9)$.
  • Why this step? We need the explicit equation of the tangent line to later find its intersection point with the parabola, which is point P.
  • Identifying parameters of the parabola: We compare the given parabola $y^2 = 4(x-9)$ with the standard form $y^2 = 4a(x-h)$.
    • From $4a = 4$, we deduce $a = 1$.
    • From $(x-h) = (x-9)$, we find $h = 9$. (This means the vertex of the parabola is at $(h,0)$, i.e., $(9,0)$).
  • Applying the tangent formula: Now, we substitute the slope of the tangent $m = -\frac{1}{2}$ (found in Step 1), and the parabola parameters $a = 1$ and $h = 9$ into the tangent equation formula: $y = m(x-h) + \frac{a}{m}$ $y = \left(-\frac{1}{2}\right)(x-9) + \frac{1}{\left(-\frac{1}{2}\right)}$ $y = -\frac{1}{2}(x-9) - 2$
  • Simplifying the equation: To eliminate fractions and express the equation in a more standard form ($Ax+By+C=0$), multiply the entire equation by 2: $2y = -(x-9) - 4$ $2y = -x + 9 - 4$ $2y = -x + 5$ Rearranging the terms: $x + 2y - 5 = 0$ or $x + 2y = 5$
  • Result: The equation of the tangent line is $x + 2y = 5$.

• Tips for Success: Accurately identifying the 'a' and 'h' values from the parabola's equation is crucial. Be careful with algebraic manipulations, especially when dealing with fractions and negative signs.

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3. Determining the Coordinates of the Point of Tangency, P

• Key Concept: The point of tangency P is the unique point where the tangent line touches the parabola. Therefore, its coordinates must satisfy both the equation of the tangent line and the equation of the parabola. We find P by solving these two equations simultaneously.

• Step-by-step Working:

  • Equations to solve:
    1. Tangent line: $x + 2y = 5$
    2. Parabola: $y^2 = 4(x-9)$
  • Why this step? Point P is the specific point on the parabola where the line touches. Finding its coordinates is essential for the final distance calculation.
  • Solving simultaneously: From the linear equation of the tangent line, it's straightforward to express $x$ in terms of $y$: $x = 5 - 2y$
  • Now, substitute this expression for $x$ into the parabola's equation: $y^2 = 4((5 - 2y) - 9)$ $y^2 = 4(5 - 2y - 9)$ $y^2 = 4(-2y - 4)$ $y^2 = -8y - 16$
  • Rearrange this into a standard quadratic equation: $y^2 + 8y + 16 = 0$
  • Solving the quadratic equation: This quadratic equation is a perfect square trinomial, which is characteristic of a tangency point (it implies a single, repeated root for the intersection). $(y+4)^2 = 0$ Solving for $y$: $y = -4$
  • Finding the corresponding x-coordinate: Substitute $y = -4$ back into the expression for $x$: $x = 5 - 2(-4)$ $x = 5 + 8$ $x = 13$
  • Result: The coordinates of the point of tangency P are $(13, -4)$.

• Tips for Success: If you do not obtain a perfect square (or a discriminant of zero) when solving the quadratic equation for the point of tangency, it's a strong indicator of an error in your tangent line equation or your algebraic calculations. This serves as a valuable self-check.

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4. Finding the Center of the Given Circle

• Key Concept: The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$. The center of this circle is at the point $(-g, -f)$.

• Step-by-step Working:

  • Given Circle Equation: We are given the equation $x^2 + y^2 - 14x - 8y + 56 = 0$.
  • Why this step? To calculate the distance from point P to the circle, we first need to know the coordinates of the circle's center.
  • Comparing with the general form: We compare the given equation with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$.
    • Comparing the coefficient of $x$: $2g = -14 \Rightarrow g = -7$.
    • Comparing the coefficient of $y$: $2f = -8 \Rightarrow f = -4$.
  • Determining the center: The center of the circle is given by $(-g, -f)$. Center $= (-(-7), -(-4)) = (7, 4)$.
  • Result: The center of the circle is $(7, 4)$.

• Tips for Success: Exercise extreme caution with signs when extracting $g$ and $f$ and then converting them to $-g$ and $-f$ for the center coordinates. A frequent mistake is to forget the negative signs required for the center's coordinates.

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5. Calculating the Distance of P from the Center of the Circle

• Key Concept: The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ in a Cartesian coordinate system is calculated using the distance formula: $D = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

• Step-by-step Working:

  • Points involved:
    • Point P: $(13, -4)$ (found in Step 3)
    • Center of the circle: $(7, 4)$ (found in Step 4)
  • Why this step? This is the final calculation that directly answers the problem's question.
  • Applying the distance formula: Let $(x_1, y_1) = (13, -4)$ and $(x_2, y_2) = (7, 4)$. $D = \sqrt{(7 - 13)^2 + (4 - (-4))^2}$ $D = \sqrt{(-6)^2 + (4 + 4)^2}$ $D = \sqrt{(-6)^2 + (8)^2}$ $D = \sqrt{36 + 64}$ $D = \sqrt{100}$ $D = 10$
  • Result: The distance of point P from the center of the circle is 10 units.

• Tips for Success: Double-check your arithmetic, especially when squaring negative numbers. Remember that $(-6)^2 = 36$, not $-36$.

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Final Answer: The distance of the point P from the centre of the circle $x^2+y^2-14 x-8 y+56=0$ is $\boxed{10}$.

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Summary and Key Takeaways

This problem serves as an excellent example of how different concepts within coordinate geometry are interconnected. To successfully solve it, we systematically applied:

1. Straight Line Properties: Using the relationship between slopes of perpendicular lines.
2. Parabola Properties: Employing the standard formula for the equation of a tangent to a parabola with a given slope.
3. Algebraic Skills: Proficiently solving simultaneous equations (a linear and a quadratic) and a quadratic equation itself.
4. Circle Properties: Extracting the center coordinates