If 2x - 4y = 9 and 6x - 12y + 7 = 0 are the t | vedclass

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(B) The given equations of the tangents are $2x - 4y - 9 = 0$ and $6x - 12y + 7 = 0$.To make the coefficients of $x$ and $y$ the same,multiply the fir

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$\frac{17}{6\sqrt{5}}$

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(B) The given equations of the tangents are $2x - 4y - 9 = 0$ and $6x - 12y + 7 = 0$.To make the coefficients of $x$ and $y$ the same,multiply the first equation by $3$: $6x - 12y - 27 = 0$.Since the tangents are parallel,the distance between them is equal to the diameter $d$ of the circle.The distance $d$ between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.Here,$A = 6, B = -12, C_1 = -27, C_2 = 7$.$d = \frac{|-27 - 7|}{\sqrt{6^2 + (-12)^2}} = \frac{|-34|}{\sqrt{36 + 144}} = \frac{34}{\sqrt{180}} = \frac{34}{6\sqrt{5}} = \frac{17}{3\sqrt{5}}$.The radius $r$ is half of the diameter: $r = \frac{d}{2} = \frac{17}{2 	imes 3\sqrt{5}} = \frac{17}{6\sqrt{5}}$.

If $2x - 4y = 9$ and $6x - 12y + 7 = 0$ are the tangents of the same circle,then its radius will be

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