The angle between the pair of tangents drawn | vedclass

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(A) The equation of the circle is $x^2+y^2-2x+4y-11=0$.Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-1, f=2, c=-11$.The radius $r = \sqrt{g^2+f^2-c}

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$\sin^{-1}\left(\frac{24}{25}\right)$

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(A) The equation of the circle is $x^2+y^2-2x+4y-11=0$.Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-1, f=2, c=-11$.The radius $r = \sqrt{g^2+f^2-c} = \sqrt{(-1)^2+2^2-(-11)} = \sqrt{1+4+11} = \sqrt{16} = 4$.The length of the tangent $L_T$ from point $(1,3)$ is $\sqrt{S_1} = \sqrt{1^2+3^2-2(1)+4(3)-11} = \sqrt{1+9-2+12-11} = \sqrt{9} = 3$.Let the angle between the pair of tangents be $2	heta$.In the right-angled triangle formed by the center,the point $(1,3)$,and the point of contact,$	an	heta = \frac{r}{L_T} = \frac{4}{3}$.We know that $\sin(2	heta) = \frac{2	an	heta}{1+	an^2	heta} = \frac{2(4/3)}{1+(4/3)^2} = \frac{8/3}{1+16/9} = \frac{8/3}{25/9} = \frac{8}{3} 	imes \frac{9}{25} = \frac{24}{25}$.Therefore,the angle $2	heta = \sin^{-1}\left(\frac{24}{25}ight)$.

The angle between the pair of tangents drawn from $(1,3)$ to the circle $x^2+y^2-2x+4y-11=0$ is

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