Question

The equation of the pair of straight lines parallel to $x$axis and touching the circle $x^2+y^2-6x-4y-12=0$

• A. $y^2–4y–21=0$
• B. $y^2+4y–21=0$
• C. $y^2–4y+21=0$
• D. $y^2+4y+21=0$

Answer 1

The correct option is A

$y^2–4y–21=0$

Explanation of the correct answer.

Step $1$: Find the radius and the center.

Given: Equation of the circle is $x^2+y^2-6x-4y-12=0$

comparing with $a{x}^2+b{y}^2+2gx+2fy+c=0$

$a=1,b=1,g=-3,f=-2,c=-12$

Since the center is $\left(-\frac{g}{a},-\frac{f}{a}\right)$ , and the radius is $\frac{1}{a}\sqrt{g^2+f^2-ac}$

Therefore,

center $=\left(3,2\right)$,

radius $=\frac{1}{1}\sqrt{{(-3)}^2+{(-2)}^2+12}$

$\Rightarrow$ radius $=5$

Step $2$: Find the equation of pair of straight line..

Since, the equation of the line parallel to $x$ axis is $y=a$.

The perpendicular distance from the line to the center is the radius.

So the distance from line $y-a=0$ to the point $\left(3,2\right)$ is

$\left|2-a\right|=5$

$\Rightarrow$$2-a=\pm 5$

$\Rightarrow$$2-a=5$

$\Rightarrow$ $a=-3$

$\Rightarrow 2-a=-5$

$\Rightarrow$ $a=7$

Therefore, the equation of pair of straight line is ,

$\left(y+3\right)\left(y-7\right)=0$

$\Rightarrow y^2-4y-21=0$

Hence, Option $\left(\mathrm{A}\right)$ is the correct answer.