Question

The equation of circle which touches the straight lines $2x-3y=10,2x-3y=4$ and its centre lies on $2x+3y=1,$ is

• A. $(x-2{)}^2+(y+1{)}^2=\frac{36}{13}$
• B. $(x-2{)}^2+(y+1{)}^2=\frac{9}{13}$
• C. $(x-3{)}^2+(y+2{)}^2=\frac{9}{13}$
• D. $(x-3{)}^2+(y+2{)}^2=\frac{36}{13}$

Answer 1

The correct option is B $(x-2{)}^2+(y+1{)}^2=\frac{9}{13}$

Given lines are $2x-3y=10$ and $2x-3y=4$
Both touching lines are parallel.
$\therefore 2r=$ distance between parallel lines
$\Rightarrow 2r=\frac{6}{\sqrt{13}}\Rightarrow r=\frac{3}{\sqrt{13}}$

The mid line between both parallel lines will pass through the centre and the equation of the line is
$2x-3y=\frac{10+4}{2}2x-3y=7\cdots \left(1\right)$

Solving $\left(1\right)$ with $2x+3y=1,$ we get centre of the circle, $C=(2,-1)$

$\therefore$ Equation of the circle is $(x-2{)}^2+(y+1{)}^2=\frac{9}{13}$