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

The circle is $$x^2 + y^2 - 2x + y - 5 = 0$$ with centre $$C = \bigl(1, -\frac{1}{2}\bigr)$$ and $$r^2 = 1 + \tfrac{1}{4} + 5 = \tfrac{25}{4}$$, so $$r = \tfrac{5}{2}$$. The tangents from $$R\bigl(\tfrac{9}{4},2\bigr)$$ touch the circle at $$P$$ and $$Q$$. To find the equation of the chord of contact $$PQ$$ we use the formula for the given circle where $$g = -1$$, $$f = \tfrac{1}{2}$$ and $$c = -5$$, namely $$ xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0. $$ Substituting $$R = \bigl(\tfrac{9}{4},2\bigr)$$ into this equation gives $$ \frac{9x}{4} + 2y - \Bigl(x + \frac{9}{4}\Bigr) + \frac{1}{2}(y + 2) - 5 = 0, $$ which simplifies to $$ \frac{5x}{4} + \frac{5y}{2} - \frac{25}{4} = 0 \;\Longrightarrow\; x + 2y - 5 = 0. $$

Substituting $$x = 5 - 2y$$ into the circle equation yields $$ (5 - 2y)^2 + y^2 - 2(5 - 2y) + y - 5 = 0 $$ or $$ 5y^2 - 15y + 10 = 0 \;\Longrightarrow\; y^2 - 3y + 2 = 0. $$ Hence $$y = 1$$ or $$y = 2$$, which gives the points $$P = (3,1)$$ and $$Q = (1,2)$$.

Using the standard coordinate formula for the area of triangle $$PQR$$ with $$P = (3,1)$$, $$Q = (1,2)$$ and $$R = \bigl(\tfrac{9}{4},2\bigr)$$, we have $$ \text{Area} = \frac{1}{2}\bigl|x_P(y_Q - y_R) + x_Q(y_R - y_P) + x_R(y_P - y_Q)\bigr| = \frac{1}{2}\bigl|3(2 - 2) + 1(2 - 1) + \tfrac{9}{4}(1 - 2)\bigr| = \frac{1}{2}\bigl|0 + 1 - \tfrac{9}{4}\bigr| = \frac{1}{2} \cdot \frac{5}{4} = \frac{5}{8}. $$

The correct answer is Option C: $$\dfrac{5}{8}$$.