The angle between the tangents drawn from the | vedclass

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(A) Given circle: $x^2+y^2+4x-6y+4=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=2, f=-3, c=4$. Centre $O = (-g, -f) = (-2, 3)$. Radius $r = \sqr

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$2 	an^{-1} \left(\frac{3}{2}ight)$

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(A) Given circle: $x^2+y^2+4x-6y+4=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=2, f=-3, c=4$. Centre $O = (-g, -f) = (-2, 3)$. Radius $r = \sqrt{g^2+f^2-c} = \sqrt{4+9-4} = \sqrt{9} = 3$. Distance from origin $P(0,0)$ to centre $O(-2,3)$ is $d = \sqrt{(-2-0)^2 + (3-0)^2} = \sqrt{4+9} = \sqrt{13}$. Let $\alpha$ be the angle between the tangent and the line joining the origin to the centre. In the right-angled triangle formed by the origin,the point of tangency,and the centre,$\sin \alpha = \frac{r}{d} = \frac{3}{\sqrt{13}}$. Then $\cos \alpha = \sqrt{1-\sin^2 \alpha} = \sqrt{1-\frac{9}{13}} = \sqrt{\frac{4}{13}} = \frac{2}{\sqrt{13}}$. Thus,$	an \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{3/\sqrt{13}}{2/\sqrt{13}} = \frac{3}{2}$. The angle between the two tangents is $2\alpha = 2 	an^{-1} \left(\frac{3}{2}ight)$.

The angle between the tangents drawn from the origin to the circle $x^2+y^2+4x-6y+4=0$ is

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