The length of the tangent drawn from the mid- | vedclass

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(D) The mid-point $P$ of the line joining the origin $(0, 0)$ and the point $(4, -4)$ is given by $P = (\frac{0+4}{2}, \frac{0-4}{2}) = (2, -2)$.The e

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$3$

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(D) The mid-point $P$ of the line joining the origin $(0, 0)$ and the point $(4, -4)$ is given by $P = (\frac{0+4}{2}, \frac{0-4}{2}) = (2, -2)$.The equation of the circle is $2x^2 + 2y^2 - y = 0$. Dividing by $2$,we get $x^2 + y^2 - \frac{1}{2}y = 0$.The length of the tangent from a point $(x_1, y_1)$ to a circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is given by $\sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c}$.Substituting $x_1 = 2, y_1 = -2, g = 0, f = -\frac{1}{4}, c = 0$:Length $= \sqrt{(2)^2 + (-2)^2 - \frac{1}{2}(-2)} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 	ext{ units.}$

The length of the tangent drawn from the mid-point of the line joining the origin and the point $(4, -4)$ to the circle $2x^2 + 2y^2 - y = 0$ is

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