Key Concepts and Formulas

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

Step-by-Step Solution

Step 1: Use the normal equation to express $\beta$ in terms of $\alpha$.

Since the center $(\alpha, \beta)$ lies on the normal $4x + 3y = 1$, we can substitute the coordinates into the equation:

$4\alpha + 3\beta = 1$

Solving for $\beta$, we get:

$\beta = \frac{1 - 4\alpha}{3} \quad (*)$

This equation will be used to reduce the number of variables.

Step 2: Set up the equidistance equation using the tangent lines.

The given tangent lines are $3x + 4y = 24$ and $3x - 4y = 32$. The distance from the center $(\alpha, \beta)$ to each tangent line must be equal to the radius $r$.

Using the distance formula:

$r = \frac{|3\alpha + 4\beta - 24|}{\sqrt{3^2 + 4^2}} = \frac{|3\alpha + 4\beta - 24|}{5}$

$r = \frac{|3\alpha - 4\beta - 32|}{\sqrt{3^2 + (-4)^2}} = \frac{|3\alpha - 4\beta - 32|}{5}$

Equating the two expressions for $r$:

$\frac{|3\alpha + 4\beta - 24|}{5} = \frac{|3\alpha - 4\beta - 32|}{5}$

$|3\alpha + 4\beta - 24| = |3\alpha - 4\beta - 32| \quad (**)$

Step 3: Substitute the expression for $\beta$ from (*) into ().**

Substitute $\beta = \frac{1 - 4\alpha}{3}$ into the equation $|3\alpha + 4\beta - 24| = |3\alpha - 4\beta - 32|$:

$|3\alpha + 4(\frac{1 - 4\alpha}{3}) - 24| = |3\alpha - 4(\frac{1 - 4\alpha}{3}) - 32|$

Multiply both sides by 3 to clear the fractions:

$|9\alpha + 4(1 - 4\alpha) - 72| = |9\alpha - 4(1 - 4\alpha) - 96|$

$|9\alpha + 4 - 16\alpha - 72| = |9\alpha - 4 + 16\alpha - 96|$

$|-7\alpha - 68| = |25\alpha - 100|$

$|7\alpha + 68| = |25\alpha - 100|$

Step 4: Solve the absolute value equation.

We have two cases:

Case 1: $7\alpha + 68 = 25\alpha - 100$ $18\alpha = 168$ $\alpha = \frac{168}{18} = \frac{28}{3}$

Case 2: $7\alpha + 68 = -(25\alpha - 100)$ $7\alpha + 68 = -25\alpha + 100$ $32\alpha = 32$ $\alpha = 1$

Step 5: Find the corresponding $\beta$ values.

For $\alpha = \frac{28}{3}$: $\beta = \frac{1 - 4(\frac{28}{3})}{3} = \frac{1 - \frac{112}{3}}{3} = \frac{-\frac{109}{3}}{3} = -\frac{109}{9}$

For $\alpha = 1$: $\beta = \frac{1 - 4(1)}{3} = \frac{-3}{3} = -1$

So the possible centers are $(\frac{28}{3}, -\frac{109}{9})$ and $(1, -1)$.

Step 6: Calculate the radius for each possible center and check the condition $r < 8$.

For $(\alpha, \beta) = (1, -1)$, using the tangent line $3x + 4y = 24$:

$r = \frac{|3(1) + 4(-1) - 24|}{5} = \frac{|3 - 4 - 24|}{5} = \frac{|-25|}{5} = 5$ Since $5 < 8$, this solution is valid.

For $(\alpha, \beta) = (\frac{28}{3}, -\frac{109}{9})$, using the tangent line $3x + 4y = 24$:

$r = \frac{|3(\frac{28}{3}) + 4(-\frac{109}{9}) - 24|}{5} = \frac{|28 - \frac{436}{9} - 24|}{5} = \frac{|4 - \frac{436}{9}|}{5} = \frac{|\frac{36 - 436}{9}|}{5} = \frac{\frac{400}{9}}{5} = \frac{80}{9} \approx 8.89$ Since $\frac{80}{9} > 8$, this solution is invalid.

Therefore, the center of the circle is $(\alpha, \beta) = (1, -1)$ and the radius is $r = 5$.

Step 7: Calculate $\alpha - \beta + r$.

$\alpha - \beta + r = 1 - (-1) + 5 = 1 + 1 + 5 = 7$

Common Mistakes & Tips

• Sign Errors: Be very careful with signs, especially when substituting and simplifying.
• Absolute Values: Remember to consider both positive and negative cases when solving absolute value equations.
• Condition r < 8: Don't forget to check the condition on the radius, as it can eliminate extraneous solutions.

Summary

We found the center and radius of the circle by using the fact that the center lies on the normal and that the radius is the perpendicular distance from the center to any tangent. By solving the resulting equations and applying the condition $r < 8$, we determined that the center is $(1, -1)$ and the radius is 5. Finally, we calculated $\alpha - \beta + r = 7$.

The final answer is $\boxed{7}$, which corresponds to option (A).