The tangents drawn from a point (2,-1) touch | vedclass

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(D) The equation of the circle is $x^2+y^2+4x-2y+1=0$. Comparing this with $x^2+y^2+2gx+2fy+c=0$,we get $g=2, f=-1, c=1$.The centre $C$ is $(-g, -f) =

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$\frac{8}{5}$

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(D) The equation of the circle is $x^2+y^2+4x-2y+1=0$. Comparing this with $x^2+y^2+2gx+2fy+c=0$,we get $g=2, f=-1, c=1$.The centre $C$ is $(-g, -f) = (-2, 1)$ and the radius $r = \sqrt{g^2+f^2-c} = \sqrt{4+1-1} = 2$.Let $P$ be the point $(2, -1)$. The distance $PC = \sqrt{(2 - (-2))^2 + (-1 - 1)^2} = \sqrt{4^2 + (-2)^2} = \sqrt{16+4} = \sqrt{20} = 2\sqrt{5}$.In the right-angled triangle $PAC$,$PA = \sqrt{PC^2 - r^2} = \sqrt{20 - 4} = \sqrt{16} = 4$.The area of triangle $PAC = \frac{1}{2} 	imes PA 	imes r = \frac{1}{2} 	imes 4 	imes 2 = 4$.The area of triangle $ABC = 2 	imes 	ext{Area}(	riangle PAC) 	imes \frac{r}{PC} = 2 	imes 4 	imes \frac{2}{2\sqrt{5}} = \frac{8}{\sqrt{5}}$.Wait,the area of $	riangle ABC$ is given by $\frac{r^3 \sqrt{PC^2-r^2}}{PC^2} = \frac{2^3 	imes 4}{20} = \frac{32}{20} = \frac{8}{5}$.

The tangents drawn from a point $(2,-1)$ touch the circle $x^2+y^2+4x-2y+1=0$ at the points $A$ and $B$. If $C$ is the centre of the circle,then the area (in sq. units) of the triangle $ABC$ is

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