Question

Two lines $4x+2y=10$ and $2x-y=20$ are touching a circle whose radius is $\sqrt{5}$ units. Then the equation of the circle which is nearest to the $x$-axis, is

• A. ${(x-6.25)}^2+{(y+2.5)}^2=5$
• B. ${(x-6.25)}^2+{(y+7.5)}^2=5$
• C. ${(x-8.75)}^2+{(y+7.5)}^2=5$
• D. ${(x-3.75)}^2+{(y+7.5)}^2=5$

Answer 1

The correct option is A ${(x-6.25)}^2+{(y+2.5)}^2=5$

Intersection point of the lines is $(6.25,-7.5)$
Equation of the angle bisector is $2x+y-5=\pm (2x-y-20)$
$\Rightarrow x=6.25$ and $y=-7.5$
To find the circle nearest to the $x$-axis, the centre of the circle should lie on $x=6.25$.
So, let the centre of the circle is $(6.25,k)$.
Then $\left|\frac{2\times 6.25+k-5}{\sqrt{5}}\right|=\sqrt{5}$
$\Rightarrow k=-2.5,-12.5$
For the circle to be nearest, $k=-2.5$
So, the equation of the circle is ${(x-6.25)}^2+{(y+2.5)}^2=5$