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(D) Let the line passing through the origin $(0, 0)$ be $y = mx$,or $mx - y = 0$.This line is a tangent to the circle $(x - 7)^2 + (y + 1)^2 = 25$ if

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$\frac{\pi}{2}$

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(D) Let the line passing through the origin $(0, 0)$ be $y = mx$,or $mx - y = 0$.This line is a tangent to the circle $(x - 7)^2 + (y + 1)^2 = 25$ if the perpendicular distance from the center $(7, -1)$ to the line is equal to the radius $r = 5$.Using the formula for the distance from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$,we have $\frac{|m(7) - (-1)|}{\sqrt{m^2 + (-1)^2}} = 5$.$\frac{|7m + 1|}{\sqrt{m^2 + 1}} = 5$.Squaring both sides: $(7m + 1)^2 = 25(m^2 + 1)$.$49m^2 + 14m + 1 = 25m^2 + 25$.$24m^2 + 14m - 24 = 0$.$12m^2 + 7m - 12 = 0$.This is a quadratic equation in $m$. Let the roots be $m_1$ and $m_2$. The product of the slopes is $m_1 m_2 = \frac{-12}{12} = -1$.Since the product of the slopes is $-1$,the two tangents are perpendicular to each other.Therefore,the angle between the two tangents is $\frac{\pi}{2}$.

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

The angle between the two tangents from the origin to the circle $(x - 7)^2 + (y + 1)^2 = 25$ is

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