The equation of the pair of straight lines parallel to x - axis and touching the circle x 2 + y 2 - 6  …

Given that: $x^2+y^2-6x-4y-12=0$

Centre of the circle: $\left(-g,-f\right)=\left(3,2\right)$

Radius of the circle: $r=\sqrt{g^2+f^2-c}=\sqrt{9+4+12}=5$

As we know that diameter is always perpendicular to the tangent.

Hence, at the ends of the diameter perpendicular to $x-\text{axis}$, tangents will be  parallel to $x-\text{axis}$.

So, tangents at $P\left(3,7\right)$and $Q=\left(3,-3\right)$ will be  parallel to $x-\text{axis}$.

Equation of line at $P\left(3,7\right)$and parallel to $x-\text{axis}$: $y=7 \Rightarrow y-7=0$

Equation of line at $Q\left(3,-3\right)$and parallel to $x-\text{axis}$: $y=-3\Rightarrow y+3=0$

Pair of straight lines:

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

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