The angle between the pair of tangents from t | vedclass

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

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(D) The equation of the circle is $x^2 + y^2 + 4x + 2y - 4 = 0$.Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = 2$,$f = 1$,and $c = -4$.The center of the circle is $C(-g, -f) = (-2, -1)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{4 + 1 - (-4)} = \sqrt{9} = 3$.The point $P$ is $(1, 1/2)$. The distance $d$ from $P$ to the center $C$ is $d = \sqrt{(1 - (-2))^2 + (1/2 - (-1))^2} = \sqrt{3^2 + (3/2)^2} = \sqrt{9 + 9/4} = \sqrt{45/4} = \frac{3\sqrt{5}}{2}$.Let $	heta$ be the angle between the tangents. Then $\sin(	heta/2) = \frac{r}{d} = \frac{3}{3\sqrt{5}/2} = \frac{2}{\sqrt{5}}$.Since $\sin(	heta/2) = \frac{2}{\sqrt{5}}$,we have $\cos(	heta/2) = \sqrt{1 - (4/5)} = 1/\sqrt{5}$.Using the double angle formula,$\cos 	heta = \cos^2(	heta/2) - \sin^2(	heta/2) = 1/5 - 4/5 = -3/5$.Thus,$	heta = \cos^{-1}(-3/5)$.

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

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