Tangents are drawn from the point (-1, -4) to | vedclass

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

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$2\sqrt{2}$ units

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(B) The equation of the circle is $x^2 + y^2 - 2x + 4y + 1 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1$,$f = 2$,and $c = 1$.Radius $R = \sqrt{g^2 + f^2 - c} = \sqrt{(-1)^2 + 2^2 - 1} = \sqrt{1 + 4 - 1} = \sqrt{4} = 2$.The length of the chord of contact from a point $(x_1, y_1)$ to a circle with radius $R$ is given by $L = \frac{2R\sqrt{S_1}}{R^2 + S_1}$,where $S_1 = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c$.For point $(-1, -4)$,$S_1 = (-1)^2 + (-4)^2 - 2(-1) + 4(-4) + 1 = 1 + 16 + 2 - 16 + 1 = 4$.Since $S_1 = 4$ and $R = 2$,the length of the chord of contact is $L = \frac{2(2)\sqrt{4}}{2^2 + 4} = \frac{4(2)}{4 + 4} = \frac{8}{8} = 1$ unit. Wait,re-evaluating: The formula for the length of the chord of contact is $\frac{2R L_t}{\sqrt{R^2 + L_t^2}}$ where $L_t$ is the length of the tangent. $L_t = \sqrt{S_1} = \sqrt{4} = 2$.$L = \frac{2 	imes 2 	imes 2}{\sqrt{2^2 + 2^2}} = \frac{8}{\sqrt{8}} = \frac{8}{2\sqrt{2}} = 2\sqrt{2}$ units.

Tangents are drawn from the point $(-1, -4)$ to the circle $x^2 + y^2 - 2x + 4y + 1 = 0$. The length of the corresponding chord of contact is:

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