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
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
\(3 / 2\)
\(3 / 4\)
\(1 / 10\)
\(1 / 20\)
Solution
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 \(\frac{3}{4}\)