If the lines 3x + 4y - 14 = 0 and 6x + 8y + 7 | vedclass

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(C) The given lines are $3x + 4y - 14 = 0$ and $6x + 8y + 7 = 0$. To make the coefficients of $x$ and $y$ the same,divide the second equation by $2$:

Vedclass · Accepted Answer

$\frac{7}{4}$

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(C) The given lines are $3x + 4y - 14 = 0$ and $6x + 8y + 7 = 0$. To make the coefficients of $x$ and $y$ the same,divide the second equation by $2$: $3x + 4y + \frac{7}{2} = 0$. Since these lines are parallel tangents to the circle,the diameter of the circle is equal to the distance between these two parallel lines. 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,$a = 3$,$b = 4$,$c_1 = -14$,and $c_2 = \frac{7}{2}$. $d = \frac{|-14 - \frac{7}{2}|}{\sqrt{3^2 + 4^2}} = \frac{|-\frac{28}{2} - \frac{7}{2}|}{\sqrt{9 + 16}} = \frac{|-\frac{35}{2}|}{5} = \frac{35}{2 	imes 5} = \frac{7}{2}$. The radius $r$ of the circle is half of the distance between the parallel tangents: $r = \frac{d}{2} = \frac{7/2}{2} = \frac{7}{4}$.

If the lines $3x + 4y - 14 = 0$ and $6x + 8y + 7 = 0$ are both tangents to a circle,then its radius is

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