Key Concepts and Formulas

• Equation of a Circle: $(x-h)^2 + (y-k)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius.
• Tangent Properties: The tangent at any point of a circle is perpendicular to the radius through the point of contact.
• Trigonometry: $\tan(2\theta) = \frac{2\tan(\theta)}{1-\tan^2(\theta)}$, Area of a triangle $= \frac{1}{2}ab\sin(C)$.

Step-by-Step Solution

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

The given equation is $x^2 + y^2 - 2x - 4y + 4 = 0$. Completing the squares, we have: $(x^2 - 2x + 1) + (y^2 - 4y + 4) - 1 - 4 + 4 = 0$ $(x - 1)^2 + (y - 2)^2 = 1$ Thus, the center of the circle is $C(1, 2)$ and the radius is $r = 1$.

Step 2: Determine $\tan(\theta/2)$.

Let $\theta$ be the angle between the tangents. We are given that $\tan(\theta) = \frac{12}{5}$. We use the identity $\tan(\theta) = \frac{2\tan(\theta/2)}{1 - \tan^2(\theta/2)}$ to find $\tan(\theta/2)$. Let $t = \tan(\theta/2)$. Then: $\frac{12}{5} = \frac{2t}{1 - t^2}$ $12(1 - t^2) = 10t$ $12 - 12t^2 = 10t$ $12t^2 + 10t - 12 = 0$ $6t^2 + 5t - 6 = 0$ $(2t + 3)(3t - 2) = 0$ Thus, $t = -\frac{3}{2}$ or $t = \frac{2}{3}$. Since $\theta \in (0, \pi)$, we have $\theta/2 \in (0, \pi/2)$, so $\tan(\theta/2)$ must be positive. Therefore, $\tan(\theta/2) = \frac{2}{3}$.

Step 3: Calculate the length of CP.

In right triangle $\Delta CAP$, we have $\tan(\theta/2) = \frac{CA}{AP} = \frac{r}{AP}$. Also, $\sin(\theta/2) = \frac{CA}{CP} = \frac{r}{CP}$. We know that $\tan(\theta/2) = \frac{2}{3}$, so we can draw a right triangle with opposite side 2 and adjacent side 3. The hypotenuse is $\sqrt{2^2 + 3^2} = \sqrt{13}$. Therefore, $\sin(\theta/2) = \frac{2}{\sqrt{13}}$. Since $CA = r = 1$, we have $\frac{1}{CP} = \frac{2}{\sqrt{13}}$, so $CP = \frac{\sqrt{13}}{2}$.

Step 4: Calculate the length of AP.

Since $\tan(\theta/2) = \frac{CA}{AP} = \frac{1}{AP} = \frac{2}{3}$, we have $AP = \frac{3}{2}$.

Step 5: Calculate the area of $\Delta PAB$.

The area of $\Delta PAB$ is given by $\frac{1}{2} \cdot PA \cdot PB \cdot \sin(\theta)$. Since $PA = PB = \frac{3}{2}$ and $\sin(\theta) = \frac{\tan(\theta)}{\sqrt{1 + \tan^2(\theta)}} = \frac{12/5}{\sqrt{1 + (12/5)^2}} = \frac{12/5}{\sqrt{169/25}} = \frac{12/5}{13/5} = \frac{12}{13}$, the area of $\Delta PAB$ is $\frac{1}{2} \cdot \frac{3}{2} \cdot \frac{3}{2} \cdot \frac{12}{13} = \frac{1}{2} \cdot \frac{9}{4} \cdot \frac{12}{13} = \frac{27}{26}$

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

In $\Delta CAB$, $CA = CB = 1$ and $\angle ACB = \pi - \theta$. Therefore, $\sin(\angle ACB) = \sin(\pi - \theta) = \sin(\theta) = \frac{12}{13}$. The area of $\Delta CAB$ is: $\frac{1}{2} \cdot CA \cdot CB \cdot \sin(\angle ACB) = \frac{1}{2} \cdot 1 \cdot 1 \cdot \frac{12}{13} = \frac{6}{13}$

Step 7: Find the ratio of the areas of $\Delta PAB$ and $\Delta CAB$.

The ratio is: $\frac{\text{Area of } \Delta PAB}{\text{Area of } \Delta CAB} = \frac{27/26}{6/13} = \frac{27}{26} \cdot \frac{13}{6} = \frac{27}{2 \cdot 6} = \frac{27}{12} = \frac{9}{4}$

Step 8: Alternative method to calculate the area of $\Delta CAB$.

Since $\sin(\theta/2) = \frac{2}{\sqrt{13}}$, $\cos(\theta/2) = \frac{3}{\sqrt{13}}$. Thus, $\sin(\theta) = 2\sin(\theta/2)\cos(\theta/2) = 2 \cdot \frac{2}{\sqrt{13}} \cdot \frac{3}{\sqrt{13}} = \frac{12}{13}$. The angle $\angle ACB = \pi - \theta$, so $\sin(\angle ACB) = \sin(\pi - \theta) = \sin(\theta) = \frac{12}{13}$. Thus, the area of $\Delta CAB$ is $\frac{1}{2} (1)(1) \sin(\angle ACB) = \frac{6}{13}$.

Step 9: Alternative method to calculate the area of $\Delta PAB$.

The area of $\Delta PAB = AP \cdot CA = \frac{1}{2}AB \cdot h$ $AB = 2 \cdot AP \sin(\theta/2) = 2 \cdot \frac{3}{2} \cdot \frac{2}{\sqrt{13}} = \frac{6}{\sqrt{13}}$ Area of $\Delta PAB = \frac{1}{2} \cdot AP \cdot BP \sin(\theta) = \frac{1}{2} (\frac{3}{2})^2 \frac{12}{13} = \frac{1}{2} \cdot \frac{9}{4} \cdot \frac{12}{13} = \frac{27}{26}$

Step 10: An even simpler approach Area($\Delta CAB$)=$\frac{1}{2}r^2\sin(\pi-\theta)=\frac{1}{2}\sin(\theta)$ Area($\Delta PAB$)=Area($\Delta CAP$) + Area($\Delta CBP$)=$2*\frac{1}{2}AP*AC=AP=r\tan(\frac{\theta}{2})=\frac{3}{2}$ $tan(\theta)=\frac{12}{5}=\frac{2tan(\frac{\theta}{2})}{1-tan^2(\frac{\theta}{2})}=\frac{2t}{1-t^2}$, where $t=tan(\frac{\theta}{2})$. $6t^2+5t-6=0$, so $t=\frac{2}{3}$ $sin(\theta)=\frac{12}{13}$. Area($\Delta CAB$)=$\frac{1}{2}sin(\theta)=\frac{6}{13}$ Area($\Delta PAB$)=$\frac{1}{2}PB*PA*sin(\theta)=\frac{1}{2}*\frac{9}{4}*\frac{12}{13}=\frac{27}{26}$ Ratio=$\frac{27/26}{6/13}=\frac{9}{4}$.

Common Mistakes & Tips

• Remember to take the positive value of $\tan(\theta/2)$ since $\theta/2$ is in the first quadrant.
• Be careful with trigonometric identities and remember the relationships between $\sin$, $\cos$, and $\tan$.
• When completing the square, ensure you subtract the added terms to maintain equality.

Summary

We first found the center and radius of the circle. Then, using the given value of $\tan(\theta)$, we calculated $\tan(\theta/2)$. Using this, we found the side lengths of the triangles $\Delta PAB$ and $\Delta CAB$. Finally, we calculated the areas of both triangles and found their ratio, which is $\frac{9}{4}$.

Final Answer

The final answer is \boxed{9/4}, which corresponds to option (B).