Question

If the lines 3x − 4y + 4 = 0 and 6x − 8y − 7 = 0 are tangents to a circle, then find the radius of the circle. [NCERT EXEMPLAR]

Answer 1

We have 3x − 4y + 4 = 0 and 6x − 8y − 7 = 0
$\Rightarrow y=\frac{3}{4}x+1\mathrm{and}y=\frac{3}{4}x-\frac{7}{8}$
Since, the slope of both the lines are equal.
Hence, the both the lines are parallel.
The distance between the parralel lines is given by
$$\begin{gathered} \left|\frac{C_1-C_2}{\sqrt{A^2+B^2}}\right| \\ =\left|\frac{4+{\displaystyle \frac{7}{2}}}{\sqrt{3^2+4^2}}\right| \\ =\left|\frac{{\displaystyle \frac{15}{2}}}{5}\right| \\ =\frac{3}{2} \end{gathered}$$​
Now, the radius is equal to the half of the distance between the parallel lines(diameter of the circle).
Hence, the radius is given by
$\frac{3}{4}$