A pair of tangents are drawn to the circle x^ | vedclass

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(D) The equation of the circle is $x^2+y^2+6x-4y-12=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=3, f=-2, c=-12$. The center $C$ is $(-g, -f) =

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$25\left(\frac{4-\pi}{4}\right)$ sq. units

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(D) The equation of the circle is $x^2+y^2+6x-4y-12=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=3, f=-2, c=-12$. The center $C$ is $(-g, -f) = (-3, 2)$ and the radius $r = \sqrt{g^2+f^2-c} = \sqrt{9+4+12} = \sqrt{25} = 5$. The distance $d$ from point $P(-4, -5)$ to the center $C(-3, 2)$ is $d = \sqrt{(-3 - (-4))^2 + (2 - (-5))^2} = \sqrt{1^2 + 7^2} = \sqrt{50} = 5\sqrt{2}$. Let $	heta$ be the angle between the tangent and the line joining the center to the point $P$. Then $\sin 	heta = \frac{r}{d} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$,so $	heta = 45^\circ$ or $\frac{\pi}{4}$ radians. The angle between the two tangents is $2\alpha$,where $\alpha = 90^\circ - 	heta = 45^\circ$. The area of the quadrilateral formed by the center,the two points of tangency,and point $P$ is $2 	imes (\frac{1}{2} 	imes r 	imes \sqrt{d^2-r^2}) = r\sqrt{d^2-r^2} = 5 	imes \sqrt{50-25} = 5 	imes 5 = 25$. The area of the two circular sectors is $2 	imes (\frac{1}{2} r^2 	heta) = r^2 	heta = 25 	imes \frac{\pi}{4}$. The area enclosed between the tangents and the circle is $25 - \frac{25\pi}{4} = 25\left(1 - \frac{\pi}{4}ight) = 25\left(\frac{4-\pi}{4}ight)$ sq. units.

$A$ pair of tangents are drawn to the circle $x^2+y^2+6x-4y-12=0$ from a point $P(-4,-5)$. The area enclosed between these tangents and the circle is:

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