Key Concepts and Formulas

• Equation of a Circle with Diameter Endpoints: The equation of a circle with diameter endpoints $P(x_1, y_1)$ and $Q(x_2, y_2)$ is given by: $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$
• Vieta's Formulas for Quadratic Equations: For a quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$:
  • Sum of roots: $\alpha + \beta = -\frac{b}{a}$
  • Product of roots: $\alpha \beta = \frac{c}{a}$

Step-by-Step Solution

1. Define Coordinates and Apply Vieta's Formulas to Abscissae

Let the points be $P(x_1, y_1)$ and $Q(x_2, y_2)$. The abscissae $x_1$ and $x_2$ are roots of $2x^2 - rx + p = 0$. Applying Vieta's formulas:

• Sum of roots: $x_1 + x_2 = -\frac{-r}{2} = \frac{r}{2}$
• Product of roots: $x_1x_2 = \frac{p}{2}$

We use Vieta's formulas to find the sum and product of the roots directly, which will be used in the circle's equation.

2. Apply Vieta's Formulas to Ordinates

The ordinates $y_1$ and $y_2$ are roots of $x^2 - sx - q = 0$. Applying Vieta's formulas:

• Sum of roots: $y_1 + y_2 = -\frac{-s}{1} = s$
• Product of roots: $y_1y_2 = \frac{-q}{1} = -q$

Again, we use Vieta's formulas to find the sum and product of the roots directly.

3. Formulate the Circle's Equation

The equation of the circle with PQ as diameter is: $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$ Expanding this equation, we get: $x^2 - (x_1 + x_2)x + x_1x_2 + y^2 - (y_1 + y_2)y + y_1y_2 = 0$ Substituting the values from steps 1 and 2, we get: $x^2 - \frac{r}{2}x + \frac{p}{2} + y^2 - sy - q = 0$ Rearranging the terms: $x^2 + y^2 - \frac{r}{2}x - sy + \frac{p}{2} - q = 0$ We use the diameter form of the circle equation because PQ is given as the diameter. Expanding it allows us to substitute the sums and products of coordinates, which we found using Vieta's formulas.

4. Compare with the Given Circle Equation

The given equation of the circle is $2(x^2 + y^2) - 11x - 14y - 22 = 0$. Dividing by 2 to match the form of our derived equation, we get: $x^2 + y^2 - \frac{11}{2}x - 7y - 11 = 0$ For a direct comparison of coefficients, it's crucial that the equations are in the same standard form. Dividing by 2 ensures that the coefficients of $x^2$ and $y^2$ are both 1.

5. Equate Coefficients and Solve for Unknowns

Comparing the coefficients of the derived equation $x^2 + y^2 - \frac{r}{2}x - sy + \frac{p}{2} - q = 0$ with the given equation $x^2 + y^2 - \frac{11}{2}x - 7y - 11 = 0$:

• Coefficient of $x$: $-\frac{r}{2} = -\frac{11}{2} \implies r = 11$
• Coefficient of $y$: $-s = -7 \implies s = 7$
• Constant term: $\frac{p}{2} - q = -11 \implies p - 2q = -22$

By equating the coefficients, we are saying that the two equations represent the same circle.

6. Calculate the Final Expression

We need to find the value of $2r + s - 2q + p$. Rearranging the terms, we get $2r + s + (p - 2q)$. Substituting the values we found:

$2(11) + 7 + (-22) = 22 + 7 - 22 = 7$

Thus, $2r + s - 2q + p = 7$.

Common Mistakes & Tips

• Sign Errors: Be very careful with signs when applying Vieta's formulas.
• Standardization: Always standardize equations before comparing coefficients.
• Algebraic Errors: Double-check all algebraic manipulations.

Summary

This problem combines geometric properties of circles with algebraic tools like Vieta's formulas. By expressing the sums and products of coordinates in terms of the given parameters and comparing the derived equation with the given equation, we efficiently solve for the unknown parameters. The final value of the expression $2r + s - 2q + p$ is 7.

The final answer is \boxed{7}.