If the tangents at the points P and Q on the circle x ^ 2 + y ^ 2 -...

Given,

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

Now plotting the diagram of the above data we have,

Now we know that,

Length of tangent is given by, $PR = QR = \sqrt{{S}_{1}}$

So, $L = \sqrt{{S}_{1}} = \sqrt{{\left(\dfrac{9}{4}\right)}^{2} + {\left(2\right)}^{2} - 2 \times \dfrac{9}{4} + 2 - 5}$

$⇒ L = \dfrac{5}{4}$

Now distance between $CR$ will be, $CR = \sqrt{{\left(\dfrac{9}{4} - 1\right)}^{2} + {\left(2 + \dfrac{1}{2}\right)}^{2}} = \sqrt{\dfrac{25}{16} + \dfrac{25}{4}} = \sqrt{\dfrac{125}{16}} = \dfrac{5\sqrt{5}}{4}$

Now finding $PQ$ by equating area, we get

$\dfrac{1}{2} \times \dfrac{5}{2} \times L = \dfrac{1}{2} \times \dfrac{PQ}{2} \times CR$

$⇒ \dfrac{1}{2} \times \dfrac{5}{2} \times \dfrac{5}{4} = \dfrac{1}{2} \times \dfrac{PQ}{2} \times \dfrac{5\sqrt{5}}{4}$

$⇒ PQ = \sqrt{5}$

Now finding area of triangle $PQR$ by heron's formula,

$s = \dfrac{2\sqrt{5} + 5}{4}$

So, area will be $∆ = \sqrt{s\left(s - a\right)\left(s - b\right)\left(s - c\right)}$

$⇒ ∆ = \sqrt{\dfrac{2\sqrt{5} + 5}{4}\left(\dfrac{2\sqrt{5} + 5}{4} - \dfrac{5}{4}\right)\left(\dfrac{2\sqrt{5} + 5}{4} - \dfrac{5}{4}\right)\left(\dfrac{2\sqrt{5} + 5}{4} - \sqrt{5}\right)}$

$⇒ ∆ = \sqrt{\dfrac{2\sqrt{5} + 5}{4}\left(\dfrac{2\sqrt{5}}{4}\right)\left(\dfrac{2\sqrt{5}}{4}\right)\left(\dfrac{5 - 2\sqrt{5}}{4}\right)}$

$⇒ ∆ = \sqrt{\left(\dfrac{5}{4}\right)\left(\dfrac{25 - 20}{16}\right)} = \dfrac{5}{8}$