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

Answer: (A)

Let the lines be $y=m_{1}x+c_1$ and $y=m_{2}x+c_2$.

Since, pair of straight lines are parallel to $x$-axis

$\therefore m_1=m_2=0$



and the lines will be $y=c_1$ and $y=c_2$.

Given circle is $x^2+y^2-6x-4y-12=0$

$\therefore$ Centre $(3,2)$ and radius $=5$

Here, the perpendicular drawn from centre to the lines are $CP$ and $C{P}^{\prime }$.

$\therefore CP=\frac{2-c_1}{\sqrt{1}}=\pm 5$

$\Rightarrow 2-c_1=\pm 5$

$\Rightarrow c_1=7$ and $c_1=-3$

Hence, the lines are

$y-7=0$, $y+3=0$

i.e., $\left(y-7\right)\left(y+3\right)=0$

$\therefore$ Pair of straight lines is

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