The angle between the pair of tangents from t | vedclass

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(B) 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 $g = 2, f = 1, c = -4$. The rad

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$\sin^{-1}(\frac{4}{5})$

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(B) 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 $g = 2, f = 1, c = -4$. The radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$. The length of the tangent from point $P(1, 1/2)$ is $\sqrt{S_1} = \sqrt{1^2 + (1/2)^2 + 4(1) + 2(1/2) - 4} = \sqrt{1 + 1/4 + 4 + 1 - 4} = \sqrt{1 + 1/4 + 1} = \sqrt{9/4} = 3/2$. Let $	heta$ be the angle between the tangents. Then $	an(	heta/2) = \frac{r}{\sqrt{S_1}} = \frac{3}{3/2} = 2$. Thus,$	heta = 2 	an^{-1}(2)$. Using the identity $2 	an^{-1}(x) = \sin^{-1}(\frac{2x}{1+x^2})$,we get $	heta = \sin^{-1}(\frac{2(2)}{1+2^2}) = \sin^{-1}(\frac{4}{5})$.

$A. (7, 2)$	$1. -4\sqrt{2}$
$B. (0, 1/\sqrt{2})$	$2. -8$
$C. (1, -1)$	$3. 4$
$D. (3, \sqrt{2})$	$4. 0$
	$5. -2\sqrt{2}$

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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$B. (0, 1/\sqrt{2})$ $2. -8$
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