If the lines 3x - 4y + 4 = 0 and 6x - 8y - 7 | vedclass

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(B) The given lines are $3x - 4y + 4 = 0$ and $6x - 8y - 7 = 0$.Dividing the second equation by $2$,we get $3x - 4y - 3.5 = 0$.Since the lines are par

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$3/4$

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(B) The given lines are $3x - 4y + 4 = 0$ and $6x - 8y - 7 = 0$.Dividing the second equation by $2$,we get $3x - 4y - 3.5 = 0$.Since the lines are parallel,the distance between them is the diameter 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,$a = 3, b = -4, c_1 = 4, c_2 = -3.5$.$d = \frac{|4 - (-3.5)|}{\sqrt{3^2 + (-4)^2}} = \frac{|7.5|}{\sqrt{9 + 16}} = \frac{7.5}{5} = 1.5$.Since the diameter is $1.5$,the radius $r = \frac{d}{2} = \frac{1.5}{2} = 0.75 = \frac{3}{4}$.

If the lines $3x - 4y + 4 = 0$ and $6x - 8y - 7 = 0$ are tangents to a circle,then the radius of the circle is

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