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(B) Let the angle between the tangents be $	heta = 	an^{-1}\left(\frac{12}{5}ight)$. Thus,$	an 	heta = \frac{12}{5}$.Since $	an 	heta = \frac{

Vedclass · Accepted Answer

$9:4$

Vedclass · Answer

(B) Let the angle between the tangents be $	heta = 	an^{-1}\left(\frac{12}{5}ight)$. Thus,$	an 	heta = \frac{12}{5}$.Since $	an 	heta = \frac{12}{5}$,we have $\sin 	heta = \frac{12}{13}$ and $\cos 	heta = \frac{5}{13}$.Let $r$ be the radius of the circle. From the equation $x^2+y^2-2x-4y+4=0$,we have $(x-1)^2 + (y-2)^2 = 1^2$,so $r=1$.Let $\alpha = 	heta/2$ be the angle between the tangent and the line joining the center to the external point $P$. Then $	an \alpha = \frac{r}{PA} = \frac{1}{PA}$.Also,$	an 	heta = \frac{2 	an \alpha}{1 - 	an^2 \alpha} = \frac{2/PA}{1 - 1/PA^2} = \frac{2 PA}{PA^2 - 1} = \frac{12}{5}$.$10 PA = 12 PA^2 - 12 \implies 6 PA^2 - 5 PA - 6 = 0 \implies (2 PA - 3)(3 PA + 2) = 0$. Since $PA > 0$,$PA = 3/2$.Area of $\Delta PAB = \frac{1}{2} (PA)^2 \sin 	heta = \frac{1}{2} \left(\frac{3}{2}ight)^2 \left(\frac{12}{13}ight) = \frac{1}{2} \cdot \frac{9}{4} \cdot \frac{12}{13} = \frac{27}{26}$.Area of $\Delta CAB = \frac{1}{2} r^2 \sin(180^\circ - 	heta) = \frac{1}{2} (1)^2 \sin 	heta = \frac{1}{2} \cdot \frac{12}{13} = \frac{6}{13}$.Ratio = $\frac{	ext{Area}(\Delta PAB)}{	ext{Area}(\Delta CAB)} = \frac{27/26}{6/13} = \frac{27}{26} \cdot \frac{13}{6} = \frac{27}{12} = \frac{9}{4}$.

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