Q89 / 160 · AP EAMCET 2018 (23 Apr Shift 2) · Mathematics
If a circle touches the lines $3 x-4 y-10=0$ and $3 x-4 y+30=0$ and its centre lies on the line $x+2 y=0$, then the equation of the circle is
- Correct answerA$x^2+y^2+4 x-2 y-11=0$
- B$x^2+y^2+2 x-4 y-11=0$
- C$x^2+y^2-4 x+2 y-11=0$
- D$x^2+y^2+2 x-y-11=0$
Solution
Distance between parallel lines
$3 x-4 y-10=0$ and $3 x-4 y+30=0$ is the length of diametre of required circle, so radius
$$
=\frac{1}{2} \frac{40}{\sqrt{9+16}}=4
$$
and mid-point of intersection of lines $3 x-4 y-10=0, x+2 y=0$ and $3 x-4 y+30=0$, $x+2 y=0$, is the centre of required circle, so centre is $\left(\frac{2-6}{2}, \frac{-1+3}{2}\right)=(-2,1)$, then equation of required circle is, $x^2+y^2+4 x-2 y-11=0$.
Hence, the correct answer is option: 'A'.