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.

Answer 1

Given: Lines $6x-8y-7=0$ …(i) and $3x-4y+4=0$ …(ii)

Multiply by $2$ in both sides of equation (ii) to make equal coefficient as in equation (i)

$\Rightarrow 2(3x-4y+4)=2\left(0\right)$

$\Rightarrow 6x-8y+8=0$ …(iii)

Clearly lines (i) and (iii) are parallel as coefficients of $x$ and $y$ are same.

Compare equation (i) with $ax+by+c_1=0$ gives $a=6,b=-8$ and $c_1=-7$

and equation (iii) with $ax+by+c_2=0$ gives $a=6,b=-8$ and $c_2=8$

Distance between two parallel lines $ax+by+c_1=0$ and $ax+by+c_2=0$

is given by $\frac{|c_1-c_2|}{\sqrt{a^2+b^2}}$

Hence distance between lines (i) and (iii) is

$\Rightarrow d=\frac{|-7-8|}{\sqrt{6^2+(-8{)}^2}}$

$\Rightarrow d=\frac{|-15|}{\sqrt{100}}$

$\Rightarrow d=\frac{15}{10}=\frac{3}{2}=$ diameter

[ Distance between parallel tangents $=$ Diameter of circle ]

Now radius of circle is $\frac{3}{4}$ units

$[\because \text{radius}=\frac{\text{diameter}}{2}]$