If the acute angle between the pair of tangen | vedclass

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(C) The equation of the circle is $S \equiv x^2+y^2-4x-8y+4=0$. The center is $(2, 4)$ and the radius $r = \sqrt{2^2+4^2-4} = \sqrt{16} = 4$. Let $d$

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$4/3$

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(C) The equation of the circle is $S \equiv x^2+y^2-4x-8y+4=0$. The center is $(2, 4)$ and the radius $r = \sqrt{2^2+4^2-4} = \sqrt{16} = 4$. Let $d$ be the distance from the origin $(0,0)$ to the center $(2,4)$,so $d = \sqrt{2^2+4^2} = \sqrt{20} = 2\sqrt{5}$. Let the angle between the tangents be $\alpha$. Then the angle between the line joining the origin to the center and one of the tangents is $\alpha/2$. In the right-angled triangle formed by the origin,the center,and the point of tangency,we have $\sin(\alpha/2) = r/d = 4/(2\sqrt{5}) = 2/\sqrt{5}$. Thus,$\cos(\alpha/2) = \sqrt{1 - \sin^2(\alpha/2)} = \sqrt{1 - 4/5} = 1/\sqrt{5}$. Therefore,$	an(\alpha/2) = \sin(\alpha/2) / \cos(\alpha/2) = (2/\sqrt{5}) / (1/\sqrt{5}) = 2$. Using the formula $	an \alpha = \frac{2 	an(\alpha/2)}{1 - 	an^2(\alpha/2)}$,we get $	an \alpha = \frac{2(2)}{1 - 2^2} = \frac{4}{1-4} = -4/3$. Since $\alpha$ is an acute angle,we take the magnitude,so $	an \alpha = 4/3$.

If the acute angle between the pair of tangents drawn from the origin to the circle $x^2+y^2-4x-8y+4=0$ is $\alpha$,then $\tan \alpha=$

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