Question

The equation of circle which touches the lines $x+2y=10,$ $x+2y=6$ and its centre lies on $2x+y=6,$ is

• A. $(3x-4{)}^2+(3y-10{)}^2=36$
• B. $(3x+4{)}^2+(3y-10{)}^2=\frac{36}{5}$
• C. $(3x-4{)}^2+(3y-10{)}^2=\frac{36}{5}$
• D. $(3x-4{)}^2+(3y-10{)}^2=\frac{144}{5}$

Answer 1

The correct option is C $(3x-4{)}^2+(3y-10{)}^2=\frac{36}{5}$

Since circle is touching both the parallel lines,
$\therefore 2r=$ distance between both the lines
$\Rightarrow 2r=\frac{4}{\sqrt{5}}\Rightarrow r=\frac{2}{\sqrt{5}}$

Now, the mid line between both the parallel lines will pass through the centre.
The equation of mid line is
$x+2y=\frac{10+6}{2}\Rightarrow x+2y=8\cdots \left(1\right)$
And $2x+y=6\cdots \left(2\right)$
Solving equation $\left(1\right)$ and $\left(2\right)$, we get
Centre $=C(\frac{4}{3},\frac{10}{3})$

Equation of circle is
${(x-\frac{4}{3})}^2+{(y-\frac{10}{3})}^2=\frac{4}{5}\Rightarrow (3x-4{)}^2+(3y-10{)}^2=\frac{36}{5}$