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

Answer 1

Answer: (A)

Let the lines be $y = m_1x + c_1$ and $y = m_2x + c_2$.

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 $CP'$.

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

$\Rightarrow 2-c_{1}=\pm 5$

$\Rightarrow c_{1}=7$ 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

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