Question

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

• 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

$y^2–4y–21=0$

Step 1: Write the equation for lines parallel to $x$ axis and find centre radius of given circle

The equation of line parallel to $x$ axis is given by $y=k$

$\Rightarrow$ $y-k=0$

For the circle $x^2+y^2+2gx+2fy+c=0$

Center is given as $\left(-g,-f\right)$ and radius $=\sqrt{g^2+f^2-c}$

On comparing given equation of circle with standard form we get,

center $=\left(3,2\right)$ and radius $=\sqrt{3^2+2^2+12}=5$

Since the lines touch the circle, the radius of the circle is equal to the perpendicular distance between the center and the line.

Step 2: Use formula for perpendicular distance between point and a line

Point $=(x_0,y_0)=(3,2)$

Line $:ax+by+c=0:y-k=0$

Distance is given by $\frac{\left|a{x}_0+b{y}_0+c\right|}{\sqrt{a^2+b^2}}$

$\Rightarrow$ $5=\frac{\left|0+1\left(2\right)-k\right|}{\sqrt{0+1}}$

$\Rightarrow$ $5=2-k$ or $5=-(2-k)$

$\Rightarrow$ $k=-3$ or $k=7$

Hence the equation of the lines are $(y-7)(y+3)=0$

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

Hence option $\left(A\right)$ is the correct answer.