Key Concepts and Formulas

• Equation of a Circle: The standard equation of a circle with center $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$.
• Tangent-Radius Property: A tangent to a circle is perpendicular to the radius at the point of tangency.
• Area of a Triangle: Area of $\Delta ABC = \frac{1}{2}ab\sin C$.

Step-by-Step Solution

Step 1: Find the center and radius of the circle.

The given equation of the circle is $x^2 + y^2 - 2x + 4y + 1 = 0$. We complete the square to rewrite the equation in standard form. $(x^2 - 2x) + (y^2 + 4y) + 1 = 0$ $(x^2 - 2x + 1) + (y^2 + 4y + 4) + 1 - 1 - 4 = 0$ $(x - 1)^2 + (y + 2)^2 = 4 = 2^2$ Thus, the center of the circle is $B(1, -2)$ and the radius is $r = 2$.

Step 2: Calculate the length of AB.

We are given that $A = (3, 1)$. We use the distance formula to find the distance between $A$ and $B$. $AB = \sqrt{(3 - 1)^2 + (1 - (-2))^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}$

Step 3: Calculate the length of AP (or AQ).

Since $AP$ is a tangent to the circle, $\Delta APB$ is a right-angled triangle with $\angle APB = 90^\circ$. Using the Pythagorean theorem in $\Delta APB$, we have $AP^2 + BP^2 = AB^2$ $AP^2 + r^2 = AB^2$ $AP^2 + 2^2 = (\sqrt{13})^2$ $AP^2 + 4 = 13$ $AP^2 = 9$ $AP = 3$ Since tangents from a point to a circle have equal length, $AP = AQ = 3$.

Step 4: Calculate $\angle ABP$.

In right-angled triangle $\Delta APB$, we have $\sin(\angle ABP) = \frac{AP}{AB}$ and $\cos(\angle ABP) = \frac{BP}{AB}$. Let $\angle ABP = \theta$. Then, $\sin \theta = \frac{AP}{AB} = \frac{2}{\sqrt{13}}$ $\cos \theta = \frac{AP}{AB} = \frac{3}{\sqrt{13}}$

Step 5: Calculate $\angle PBQ$.

Since $BP$ and $BQ$ are radii of the circle, $BP = BQ$. Also, $AP = AQ$. Thus, $AB$ bisects $\angle PBQ$, and $\angle PBA = \angle QBA = \theta$. Therefore, $\angle PBQ = 2\theta$.

Step 6: Calculate the area of $\Delta APQ$.

The area of $\Delta APQ$ is given by $\text{Area}(\Delta APQ) = \frac{1}{2} AP \cdot AQ \cdot \sin(\angle PAQ)$ Since $\angle PAB = \angle QAB$, we have $\angle PAQ = 2 \angle PAB$. Also, $\angle PAB = 90^{\circ} - \theta$. Therefore $\angle PAQ = 2(90^\circ - \theta) = 180^\circ - 2\theta$. $\sin(\angle PAQ) = \sin(180^\circ - 2\theta) = \sin(2\theta) = 2 \sin \theta \cos \theta = 2 \cdot \frac{2}{\sqrt{13}} \cdot \frac{3}{\sqrt{13}} = \frac{12}{13}$ $\text{Area}(\Delta APQ) = \frac{1}{2} \cdot 3 \cdot 3 \cdot \frac{12}{13} = \frac{54}{13}$

Step 7: Calculate the area of $\Delta BPQ$.

The area of $\Delta BPQ$ is given by $\text{Area}(\Delta BPQ) = \frac{1}{2} BP \cdot BQ \cdot \sin(\angle PBQ) = \frac{1}{2} r^2 \sin(2\theta) = \frac{1}{2} (2^2) \cdot \frac{12}{13} = \frac{24}{13}$

Step 8: Calculate the ratio of the areas.

$\frac{\text{Area}(\Delta APQ)}{\text{Area}(\Delta BPQ)} = \frac{54/13}{24/13} = \frac{54}{24} = \frac{9}{4}$

Step 9: Calculate the final expression.

We need to find the value of $8 \cdot \frac{\text{Area}(\Delta APQ)}{\text{Area}(\Delta BPQ)}$. $8 \cdot \frac{\text{Area}(\Delta APQ)}{\text{Area}(\Delta BPQ)} = 8 \cdot \frac{9}{4} = 2 \cdot 9 = 18$

Common Mistakes & Tips

• Be careful with trigonometric identities, especially when dealing with double angles.
• Remember that tangents from an external point to a circle are equal in length.
• Drawing a diagram is very helpful in visualizing the problem.

Summary

We found the center and radius of the circle by completing the square. Then we calculated the length of $AB$, $AP$, and used trigonometric relationships to find the areas of triangles $APQ$ and $BPQ$. Finally, we calculated the required expression $8 \cdot \frac{\text{Area}(\Delta APQ)}{\text{Area}(\Delta BPQ)}$, which equals 18.

The final answer is \boxed{18}.