The area of the triangle (in sq. units) forme | vedclass

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(A) The equation of the circle is $x^2+y^2-2x-2y-7=0$. The center is $C(1,1)$ and the radius $r = \sqrt{1^2+1^2-(-7)} = \sqrt{9} = 3$.The distance fro

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$4.5$

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(A) The equation of the circle is $x^2+y^2-2x-2y-7=0$. The center is $C(1,1)$ and the radius $r = \sqrt{1^2+1^2-(-7)} = \sqrt{9} = 3$.The distance from $P(4,4)$ to $C(1,1)$ is $d = \sqrt{(4-1)^2+(4-1)^2} = \sqrt{3^2+3^2} = 3\sqrt{2}$.The length of the tangent $L = \sqrt{d^2-r^2} = \sqrt{(3\sqrt{2})^2 - 3^2} = \sqrt{18-9} = 3$.The chord of contact is $T=0$,given by $x(4)+y(4)-(x+4)-(y+4)-7=0$,which simplifies to $3x+3y-15=0$ or $x+y=5$.The distance from $C(1,1)$ to the chord $x+y-5=0$ is $h = \frac{|1+1-5|}{\sqrt{1^2+1^2}} = \frac{3}{\sqrt{2}}$.The length of the chord of contact is $2\sqrt{r^2-h^2} = 2\sqrt{9 - \frac{9}{2}} = 2\sqrt{\frac{9}{2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2}$.The area of the triangle formed by the two tangents and the chord of contact is given by $\frac{RL^3}{R^2+L^2}$,where $R$ is the radius of the circle and $L$ is the length of the tangent. Here $R=3$ and $L=3$,so Area $= \frac{3 	imes 3^3}{3^2+3^2} = \frac{81}{18} = 4.5$ sq. units.

The area of the triangle (in sq. units) formed by the tangents drawn from $P(4,4)$ to the circle $S \equiv x^2+y^2-2x-2y-7=0$ and the chord of contact of $P$ with respect to $S=0$ is

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