Two tangents are drawn from a point P to the circle x ^ 2 + y ^ 2 -...

Let, the angle between the tangents $PA$ and $PB$ from the point $P$ to the circle ${x}^{2} + {y}^{2} - 2x - 4y + 4 = 0$ with centre $C$ is $\theta .$

Given $\theta = \tan^{- 1}\left(\dfrac{12}{5}\right) ⇒ tan\theta = \dfrac{12}{5} .$

We know that the centre and radius of a circle ${x}^{2} + {y}^{2} + 2gx + 2fy + c = 0$ are respectively \left(- g, - f\right)$and$\sqrt{{g}^{2} + {f}^{2} - c} .$</p><p>Thus, the centre and radius of the circle${x}^{2} + {y}^{2} - 2x - 4y + 4 = 0$are respectively \left(1, 2\right)$ and $\sqrt{{1}^{2} + {2}^{2} - 4} = 1$ unit.

We know that the line joining the centre of a circle to the point of contact of the tangent is perpendicular to the tangent, hence $∠ CAP = ∠ CBP = 90 °$ also $CP$ bisects the angle $∠ APB$ as $∆ CAP ≅ ∆ CBP$ by $SSS$ congruency.

Thus, we have $∠ CPA = ∠ CPB = \dfrac{\theta}{2}$

In $∆ CAP,$ we have $\tan \left(\dfrac{\theta}{2}\right) = \dfrac{AC}{AP}$

$⇒ \tan \left(\dfrac{\theta}{2}\right) = \dfrac{1}{AP}$

$⇒ AP = \cot \left(\dfrac{\theta}{2}\right)$

Now, area of $\delta PAB = \dfrac{1}{2}\left(PA\right)\left(PB\right)sin\theta$

Since, tangents to a circle from an external point are equal, hence $PA = PB$

Area of $\delta PAB = \dfrac{1}{2}{\left(PA\right)}^{2}sin\theta$

$= \dfrac{1}{2}co{t}^{2}\left(\dfrac{\theta}{2}\right)sin\theta$

$= \dfrac{1}{2}\dfrac{\cos^{2}\left(\dfrac{\theta}{2}\right)}{\sin^{2}\left(\dfrac{\theta}{2}\right)}sin\theta$

Using $\cos 2x = 2\cos^{2}x - 1 = 1 - 2\sin^{2}x,$ we get

Area of $\delta PAB = \dfrac{1}{2}\left(\dfrac{1 + cos\theta}{1 - cos\theta}\right)sin\theta$

We have $tan\theta = \dfrac{12}{5} ⇒ sin\theta = \dfrac{12}{13} \text{ and } cos\theta = \dfrac{5}{13}$

Area of $\delta PAB = \dfrac{1}{2}\left(\dfrac{1 + \dfrac{5}{13}}{1 - \dfrac{5}{13}}\right)\left(\dfrac{12}{13}\right)$

$= \dfrac{1}{2} \times \dfrac{18}{8} \times \dfrac{12}{13} = \dfrac{27}{26}$ sq units.

And, area of $\delta CAB = \dfrac{1}{2}\left(AC\right)\left(BC\right)\sin \left(\pi - \theta\right)$

$= \dfrac{1}{2}sin\theta = \dfrac{1}{2}\left(\dfrac{12}{13}\right) = \dfrac{6}{13}$ sq units.

$∴\dfrac{areaof\delta PAB}{areaof\delta CAB} = \dfrac{\left(\dfrac{27}{26}\right)}{\left(\dfrac{6}{13}\right)} = \dfrac{9}{4} .$