Key Concepts and Formulas

• Normal to a Circle: A line normal to a circle passes through the center of the circle. The intersection of two normals gives the center.
• Tangent to a Circle: The distance from the center of the circle to a tangent line is equal to the radius.
• Distance from a Point to a Line: The distance $d$ from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is given by $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.

Step-by-Step Solution

Step 1: Finding the Center of the Circle $(h, k)$

The center of the circle is the intersection of the two normal lines. We solve the system of equations: $y + 2x = \sqrt{11} + 7\sqrt{7} \quad (1)$ $2y + x = 2\sqrt{11} + 6\sqrt{7} \quad (2)$ Multiply equation (2) by 2: $4y + 2x = 4\sqrt{11} + 12\sqrt{7} \quad (3)$ Subtract equation (1) from equation (3): $(4y + 2x) - (y + 2x) = (4\sqrt{11} + 12\sqrt{7}) - (\sqrt{11} + 7\sqrt{7})$ $3y = 3\sqrt{11} + 5\sqrt{7}$ $y = \sqrt{11} + \frac{5}{3}\sqrt{7}$ Substitute the value of $y$ back into equation (2): $2(\sqrt{11} + \frac{5}{3}\sqrt{7}) + x = 2\sqrt{11} + 6\sqrt{7}$ $2\sqrt{11} + \frac{10}{3}\sqrt{7} + x = 2\sqrt{11} + 6\sqrt{7}$ $x = 6\sqrt{7} - \frac{10}{3}\sqrt{7} = \frac{18\sqrt{7} - 10\sqrt{7}}{3} = \frac{8}{3}\sqrt{7}$ Thus, the center of the circle is $(h, k) = \left(\frac{8}{3}\sqrt{7}, \sqrt{11} + \frac{5}{3}\sqrt{7}\right)$.

Step 2: Finding the Radius of the Circle $r$

The radius is the distance from the center to the tangent line. The tangent line is given by: $\sqrt{11}y - 3x = \frac{5\sqrt{77}}{3} + 11$ Rewrite the equation in the form $Ax + By + C = 0$: $3x - \sqrt{11}y + \frac{5\sqrt{77}}{3} + 11 = 0$ So, $A = 3$, $B = -\sqrt{11}$, and $C = \frac{5\sqrt{77}}{3} + 11$. We use the distance formula: $r = \frac{|3(\frac{8}{3}\sqrt{7}) - \sqrt{11}(\sqrt{11} + \frac{5}{3}\sqrt{7}) + \frac{5\sqrt{77}}{3} + 11|}{\sqrt{3^2 + (-\sqrt{11})^2}}$ $r = \frac{|8\sqrt{7} - 11 - \frac{5\sqrt{77}}{3} + \frac{5\sqrt{77}}{3} + 11|}{\sqrt{9 + 11}}$ $r = \frac{|8\sqrt{7}|}{\sqrt{20}} = \frac{8\sqrt{7}}{2\sqrt{5}} = \frac{4\sqrt{7}}{\sqrt{5}} = \frac{4\sqrt{35}}{5}$ Therefore, $r^2 = \left(\frac{4\sqrt{35}}{5}\right)^2 = \frac{16 \cdot 35}{25} = \frac{16 \cdot 7}{5} = \frac{112}{5}$.

Step 3: Calculating the Final Expression

We want to find $(5h - 8k)^2 + 5r^2$. First, calculate $5h - 8k$: $5h = 5(\frac{8}{3}\sqrt{7}) = \frac{40}{3}\sqrt{7}$ $8k = 8(\sqrt{11} + \frac{5}{3}\sqrt{7}) = 8\sqrt{11} + \frac{40}{3}\sqrt{7}$ $5h - 8k = \frac{40}{3}\sqrt{7} - (8\sqrt{11} + \frac{40}{3}\sqrt{7}) = -8\sqrt{11}$ $(5h - 8k)^2 = (-8\sqrt{11})^2 = 64 \cdot 11 = 704$ Next, calculate $5r^2$: $5r^2 = 5(\frac{112}{5}) = 112$ Finally, $(5h - 8k)^2 + 5r^2 = 704 + 112 = 816$

Common Mistakes & Tips

• Be careful with signs when substituting values into the distance formula.
• Double-check the arithmetic when simplifying radical expressions and fractions.
• Remember that $(\sqrt{a})^2 = a$, but $\sqrt{a^2} = |a|$.

Summary

We found the center of the circle by solving the system of equations formed by the normal lines. Then, we found the radius using the point-to-line distance formula with the tangent line and the center. Finally, we substituted these values into the expression $(5h - 8k)^2 + 5r^2$ to get the final answer.

The final answer is $\boxed{816}$.