The equation of pair of straight lines parallel to $X$-axis and touching the circle $x^2+y^2-6 x-4 y-12=0$
The equation of pair of straight lines parallel to $X$-axis and touching the circle $x^2+y^2-6 x-4 y-12=0$
$y^2-4 y-21=0$
$y^2+4 y-21=0$
$y^2-4 y+21=0$
$y^2+4 y+21=0$
Solution
Given circle is $x^2+y^2-6 x-4 y-12=0$
$
\begin{aligned}
\text { Centre } & =(3,2) \\
\text { Radius } & =\sqrt{(3)^2+(2)^2+12}=\sqrt{9+4+12}=\sqrt{25} \\
r & =5
\end{aligned}
$
Since, tangents are paralle to $X$-axis
$\therefore$ Let $y=k$ be the tangent to the given circle
$\Rightarrow y-k=0$ touches the circle
$\therefore$ Perpendicular distance
from $C(3,2)$ to $y-k=0=$ radius
$
\frac{|2-k|}{\sqrt{1}}=5
$
$
\begin{aligned}
|2-k| & =5 \\
2-k & = \pm 5 \\
2-k=5 \text { (or) } 2-k & =-5 \\
k=-3 \text { (or) } k & =7
\end{aligned}
$
$\therefore$ Required equation of tangents are
$
y=-3 \text { and } y=7
$
$\Rightarrow(y+3)=0$ and $(y-7)=0$
$\therefore$ Equation of pair of tangents is $(y+3)(y-7)=0$
$
y^2-4 y-21=0
$
Hence, option (1) is correct