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If the lines $3 x-4 y+4=0$ and $6 x-8 y-7=0$ are tangents to a circle, then radius of the circle is
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The correct answer is:
$\frac{3}{4}$
The diameter of the circle is perpendicular distance between the parallel lines (tangents) $3 x-4 y+4=0$ and
$3 x-4 y-\frac{7}{2}=0$ and so it is equal to $\frac{4}{\sqrt{9+16}}+\frac{7 / 2}{\sqrt{9+16}}=\frac{3}{2}$. Hence radius is $\frac{3}{4}$.
$3 x-4 y-\frac{7}{2}=0$ and so it is equal to $\frac{4}{\sqrt{9+16}}+\frac{7 / 2}{\sqrt{9+16}}=\frac{3}{2}$. Hence radius is $\frac{3}{4}$.
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