If the chord of contact of the point P(1, 1) | vedclass

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(B) The equation of the circle is $S = x^2 + y^2 + 4x + 6y - 3 = 0$. The center $O$ is $(-2, -3)$ and the radius $r = \sqrt{(-2)^2 + (-3)^2 - (-3)} =

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$\frac{108}{25}$

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(B) The equation of the circle is $S = x^2 + y^2 + 4x + 6y - 3 = 0$. The center $O$ is $(-2, -3)$ and the radius $r = \sqrt{(-2)^2 + (-3)^2 - (-3)} = \sqrt{4 + 9 + 3} = 4$.For the point $P(1, 1)$,the distance $OP = \sqrt{(1 - (-2))^2 + (1 - (-3))^2} = \sqrt{3^2 + 4^2} = 5$.Let $A$ and $B$ be the points of contact. The length of the tangent $PA = \sqrt{OP^2 - r^2} = \sqrt{5^2 - 4^2} = 3$.Let $\angle AOP = 	heta$. In $	riangle OAP$,$\sin 	heta = \frac{PA}{OP} = \frac{3}{5}$ and $\cos 	heta = \frac{OA}{OP} = \frac{4}{5}$.The length of the chord of contact $AB = 2 	imes PA \sin 	heta = 2 	imes 3 	imes \frac{3}{5} = \frac{18}{5}$.The distance from $O$ to the chord $AB$ is $OQ = OA \cos 	heta = 4 	imes \frac{4}{5} = \frac{16}{5}$.The distance $PQ = OP - OQ = 5 - \frac{16}{5} = \frac{9}{5}$.The area of $	riangle PAB = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes AB 	imes PQ = \frac{1}{2} 	imes \frac{18}{5} 	imes \frac{9}{5} = \frac{81}{25}$.Wait,re-evaluating: The area of $	riangle PAB = \frac{1}{2} 	imes AB 	imes PQ = \frac{1}{2} 	imes \frac{18}{5} 	imes \frac{9}{5} = \frac{81}{25}$.Let's re-calculate: $Area = \frac{r \cdot L^3}{r^2 + L^2} = \frac{4 \cdot 3^3}{4^2 + 3^2} = \frac{4 \cdot 27}{25} = \frac{108}{25}$.

If the chord of contact of the point $P(1, 1)$ with respect to the circle $S = x^2 + y^2 + 4x + 6y - 3 = 0$ meets the circle $S = 0$ at $A$ and $B$,then the area of $\triangle PAB$ is

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