First slide

Definition of a circle

Question

$T w o \tan g e n t s a r e d r a w n f r o m a p o i n t P t o t h e c i r c l e$ $x^{2} + y^{2} - 2 x - 4 y + 4 = 0$ , $s u c h t h a t t h e a n g l e b e t w e e n t h e s e \tan g e n t s i s$ $\tan^{- 1} (\frac{12}{5})$ , $w h e r e$ $\tan^{- 1} (\frac{12}{5}) \in (0, π)$ . $I f t h e c e n t e r o f t h e a r e a s o f$ $Δ P A B a n d Δ C A B$ is

Moderate

A

9:4
B

2:1
C

3:1
D

11:4

Solution

$\begin{array}{l} The given circle is x^{2} + y^{2} - 2 x - 4 y + 4 = 0 \\ Its Center (1, 2) and radius is r = 1 unit \\ Given angle between the tangents drawn from P {to the circle is Tan}^{- 1} (\frac{12}{5}) \\ 2 θ = \tan^{- 1} (\frac{12}{5}) \\ \frac{2 \tan θ}{1 - \tan^{2} θ} = \frac{12}{5} \\ cross multiply and then solve the quadratic equation \\ \tan θ = \frac{2}{3} \\ In triangle APC, tan θ = \frac{r}{A P} \\ A P = \frac{3}{2} \\ Hence the ratios of the areas of triangles PAB and CAB is \frac{\frac{1}{2} \cdot P A \cdot P B \cdot \sin P}{\frac{1}{2} \cdot C A \cdot C B \cdot \sin C} \\ Here both angles P and C are complimentary \\ Hence the ratio is \end{array}$
$∴ \frac{A r (Δ P A B)}{A r (Δ C A B)} = \frac{P A . P B . \sin P}{A C . B C \sin C} = \frac{\frac{3}{2} . \frac{3}{2} . \frac{12}{13}}{1.1. \frac{12}{13}} = \frac{9}{4} = 9 : 4$

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