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(D) The equation of the circle is $x^2 + y^2 - 2x + y - 5 = 0$.Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1$,$f = \frac{1}{2}$,and $c

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$\frac{5}{8}$

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(D) The equation of the circle is $x^2 + y^2 - 2x + y - 5 = 0$.Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1$,$f = \frac{1}{2}$,and $c = -5$.The center of the circle is $C(-g, -f) = (1, -\frac{1}{2})$.The radius $r$ is $\sqrt{g^2 + f^2 - c} = \sqrt{1 + \frac{1}{4} + 5} = \sqrt{\frac{25}{4}} = \frac{5}{2}$.Let $L$ be the length of the tangent $PR = QR$. $L = \sqrt{S_1} = \sqrt{(\frac{9}{4})^2 + 2^2 - 2(\frac{9}{4}) + 2 - 5} = \sqrt{\frac{81}{16} + 4 - \frac{9}{2} + 2 - 5} = \sqrt{\frac{81 - 72 + 16}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.The distance $d$ from the center $C$ to the point $R$ is $d = \sqrt{(\frac{9}{4} - 1)^2 + (2 + \frac{1}{2})^2} = \sqrt{(\frac{5}{4})^2 + (\frac{5}{2})^2} = \sqrt{\frac{25}{16} + \frac{25}{4}} = \sqrt{\frac{125}{16}} = \frac{5\sqrt{5}}{4}$.The area of $	riangle PQR$ is given by $\frac{r L^3}{r^2 + L^2} = \frac{(\frac{5}{2}) \cdot (\frac{5}{4})^3}{(\frac{5}{2})^2 + (\frac{5}{4})^2} = \frac{\frac{5}{2} \cdot \frac{125}{64}}{\frac{25}{4} + \frac{25}{16}} = \frac{\frac{625}{128}}{\frac{100 + 25}{16}} = \frac{625}{128} \cdot \frac{16}{125} = \frac{5}{8}$.

If the tangents at the points $P$ and $Q$ on the circle $x^2 + y^2 - 2x + y = 5$ meet at the point $R \left(\frac{9}{4}, 2\right)$,then the area of the triangle $PQR$ is

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