Key Concepts and Formulas

• Equation of a Circle: The general equation of a circle is $(x-h)^2 + (y-k)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius.
• Length of Tangent: The length of the tangent from an external point $(x_1, y_1)$ to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is given by $\sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c}$.
• Area of Triangle PQR: If $r$ is the radius of the circle and $l$ is the length of the tangent from the external point $R$ to the circle, then the area of $\triangle PQR$ is given by $\frac{rl^3}{r^2 + l^2}$.

Step-by-Step Solution

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

The equation of the circle is given by $x^2 + y^2 - 2x + y = 5$. We need to rewrite this in the standard form $(x-h)^2 + (y-k)^2 = r^2$ to find the center $(h, k)$ and radius $r$.

Completing the square for the $x$ terms: $x^2 - 2x = (x - 1)^2 - 1$ Completing the square for the $y$ terms: $y^2 + y = (y + \frac{1}{2})^2 - \frac{1}{4}$

Substituting these back into the equation: $(x - 1)^2 - 1 + (y + \frac{1}{2})^2 - \frac{1}{4} = 5$ $(x - 1)^2 + (y + \frac{1}{2})^2 = 5 + 1 + \frac{1}{4} = \frac{25}{4}$

Therefore, the center of the circle is $(1, -\frac{1}{2})$ and the radius is $r = \sqrt{\frac{25}{4}} = \frac{5}{2}$.

Step 2: Find the length of the tangent from point R to the circle.

The point $R$ is given as $(\frac{9}{4}, 2)$. The equation of the circle is $x^2 + y^2 - 2x + y - 5 = 0$. The length of the tangent $l$ from point $R$ to the circle is given by:

$l = \sqrt{(\frac{9}{4})^2 + (2)^2 - 2(\frac{9}{4}) + (2) - 5}$ $l = \sqrt{\frac{81}{16} + 4 - \frac{18}{4} + 2 - 5}$ $l = \sqrt{\frac{81}{16} - \frac{72}{16} + 1}$ $l = \sqrt{\frac{9}{16} + 1}$ $l = \sqrt{\frac{25}{16}} = \frac{5}{4}$

Step 3: Calculate the area of triangle PQR.

Now we use the formula for the area of $\triangle PQR$: Area $= \frac{rl^3}{r^2 + l^2}$

We have $r = \frac{5}{2}$ and $l = \frac{5}{4}$. Substituting these values:

Area $= \frac{(\frac{5}{2})(\frac{5}{4})^3}{(\frac{5}{2})^2 + (\frac{5}{4})^2}$ Area $= \frac{(\frac{5}{2})(\frac{125}{64})}{\frac{25}{4} + \frac{25}{16}}$ Area $= \frac{\frac{625}{128}}{\frac{100}{16} + \frac{25}{16}}$ Area $= \frac{\frac{625}{128}}{\frac{125}{16}}$ Area $= \frac{625}{128} \cdot \frac{16}{125}$ Area $= \frac{5}{8} \cdot \frac{1}{1} = \frac{5}{8}$

Step 4: Verify the answer.

The area of triangle PQR is $\frac{5}{8}$.

Common Mistakes & Tips

• Completing the Square: Be careful when completing the square to correctly identify the center and radius of the circle. A common mistake is to forget to subtract the squared term when completing the square.
• Tangent Length Formula: Ensure you use the correct formula for the length of the tangent from an external point.
• Area Formula: Remember the formula for the area of triangle PQR in terms of the radius and tangent length. Memorizing this formula saves time.

Summary

We first found the center and radius of the given circle by completing the square. Then, we calculated the length of the tangent from the external point R to the circle. Finally, we used the formula for the area of the triangle formed by the tangents and the chord of contact to arrive at the area of triangle PQR, which is $\frac{5}{8}$.

Final Answer

The final answer is \boxed{\frac{5}{8}}, which corresponds to option (B).