Let the tangents drawn from P(-1, -1) to the | vedclass

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(A) The equation of the circle is $x^2 + y^2 - 2x - 4y - 4 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get the center $C(1, 2)$ and radius

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$\frac{24}{13}$

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(A) The equation of the circle is $x^2 + y^2 - 2x - 4y - 4 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get the center $C(1, 2)$ and radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{1^2 + 2^2 - (-4)} = \sqrt{9} = 3$. The distance $PC$ from $P(-1, -1)$ to $C(1, 2)$ is $d = \sqrt{(1 - (-1))^2 + (2 - (-1))^2} = \sqrt{2^2 + 3^2} = \sqrt{13}$. The length of the tangent $L = \sqrt{d^2 - r^2} = \sqrt{13 - 9} = \sqrt{4} = 2$. Let $h$ be the altitude from $A$ to $PC$ and $k$ be the length $PM$ where $M$ is the intersection of $AB$ and $PC$. In $	riangle PAC$,$\angle PAC = 90^\circ$. The area of $	riangle PAC = \frac{1}{2} 	imes AC 	imes AP = \frac{1}{2} 	imes 3 	imes 2 = 3$. Also,area of $	riangle PAC = \frac{1}{2} 	imes PC 	imes h = \frac{1}{2} 	imes \sqrt{13} 	imes h = 3$,so $h = \frac{6}{\sqrt{13}}$. In $	riangle PAM$,$PM = \sqrt{AP^2 - h^2} = \sqrt{4 - \frac{36}{13}} = \sqrt{\frac{52 - 36}{13}} = \sqrt{\frac{16}{13}} = \frac{4}{\sqrt{13}}$. The length of the chord $AB = 2h = \frac{12}{\sqrt{13}}$. The area of $	riangle PAB = \frac{1}{2} 	imes AB 	imes PM = \frac{1}{2} 	imes \frac{12}{\sqrt{13}} 	imes \frac{4}{\sqrt{13}} = \frac{24}{13}$.

Let the tangents drawn from $P(-1, -1)$ to the circle $x^2 + y^2 - 2x - 4y - 4 = 0$ touch the circle at the points $A$ and $B$. Then the area of the triangle $PAB$ (in square units) is

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