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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 the radius of the circle is
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The correct answer is:
\(3 / 4\)
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 \(\frac{3}{4}\)
\(\frac{4}{\sqrt{9+16}}+\frac{7 / 2}{\sqrt{9+16}}=\frac{3}{2}\). Hence radius is \(\frac{3}{4}\)
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