The radius of the circle having $3 x-4 y+4=0$ and $6 x-8 y-7=0$ as its tangents is
The radius of the circle having $3 x-4 y+4=0$ and $6 x-8 y-7=0$ as its tangents is
$\frac{3}{2}$
$3$
$6$
$\frac{3}{4}$
Solution
Equation of the given tangents are
$\begin{aligned} & E_1=3 x-4 y+4=0 \\ & E_2=6 x-8 y-7=0\end{aligned}$
$\Rightarrow \quad 3 x-4 y-\frac{7}{2}=0$
Here, $a=3, b=-4 c_1=4, c_2=\frac{-7}{2}$
Since, slopes of the given tangents are equal, i.e. $\frac{3}{4}$.
$\therefore$ The given tangents are parallel,
$d=\left|\frac{c_2-c_1}{\sqrt{a^2+b^2}}\right|=\left|\frac{4-\left(\frac{-7}{2}\right)}{\sqrt{3^2+(-4)^2}}\right|$
$d=\frac{\frac{15}{2}}{\sqrt{9+16}}=\frac{15}{2 \times 5}=\frac{3}{2}$
As we know that distance between two parallel tangents of a circle is equal to the diameter of that circle.
$\therefore$ Diameter of given circle $=\frac{3}{2}$
$\therefore$ Radius of the given circle $=\frac{\text { Diameter }}{2}$
$=\left(\frac{3 / 2}{2}\right)=\frac{3}{4}$