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.

• A. $\frac{3}{4}$
• B. $\frac{4}{3}$
• C. $\frac{1}{4}$
• D. $\frac{7}{4}$

Answer 1

The correct option is A $\frac{3}{4}$

Equation of the given tangents are-

$t_1:3x-4y+4=0$

$t_2:6x-8y-7=0\Rightarrow 3x-4y-\frac{7}{2}=0$

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.

As we know that, the distance between two parallel lines given by,

$d=\frac{|c_2-c_1|}{\sqrt{a^2+b^2}}$

$\therefore$ distance between $t_1\&t_2=\frac{|4-(-\frac{7}{2})|}{\sqrt{3^2+{(-4)}^2}}=\frac{\left(\frac{15}{2}\right)}{\sqrt{9+16}}=\frac{15}{2\times 5}=\frac{3}{2}$

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 $=\frac{diameter}{2}=\frac{\left(\frac{3}{2}\right)}{2}=\frac{3}{4}$

Hence, the radius of the given circle is $\frac{3}{4}$.