Since you know the center and the point of tangency, you can compute the slope of the radius to the point of tangency. Since the center is $(3, 0)$ and the point of tangency is $(2, 2\sqrt{2})$, the slope of the radius to the point of tangency is
$$m_r = \frac{2\sqrt{2} - 0}{3 - 2} = 2\sqrt{2}$$
The slope of the tangent line to the circle is perpendicular to the radius at the point of tangency. If two non-vertical lines are perpendicular, their slopes are negative reciprocals, so the slope of the tangent line to the circle at $(2, 2\sqrt{2})$ is the negative reciprocal of the slope of the radius to the point of tangency. Thus, the tangent line has slope
$$m_{\perp} = -\frac{1}{2\sqrt{2}} = -\frac{1}{2\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = -\frac{\sqrt{2}}{4}$$
You can then use the point-slope equation
$$y - y_0 = m(x - x_0)$$
to write the equation of the tangent line, where $(x_0, y_0)$ is the point $(2, 2\sqrt{2})$ and $m$ is the slope of the tangent line. The equation of the tangent line to the circle $(x - 3)^2 + y^2 = 9$ at $(2, 2\sqrt{2})$ is
$$y - 2\sqrt{2} = -\frac{\sqrt{2}}{4}(x - 2)$$
A reflection in the $x$-axis sends point $(x, y)$ to the point $(x, -y)$. Thus, the reflection of the point $(2, 2\sqrt{2})$ in the $x$-axis is $(2, -2\sqrt{2})$. To find the equation of the tangent line to the circle at this point, follow the steps outlined above with $(2, -2\sqrt{2})$ replacing $(2, 2\sqrt{2})$.
Calculate Power of the circle by shifting P to origin and the circle also accordingly for same relative position.
$$ ( x + 5 + 8)^2 + ( y -9 -7)^2 = 25 $$
Tangent length is square root of Power = $ \sqrt{13^2 + 16^2 -25} =20. $ by the Pole/Polar method.
Best Answer
The line is tangent to the circle iff they intersect at exactly one point. The intersection is given by a quadratic equation, the number of solutions is given by the sign of the discriminant (you want the discriminant to be $0$).
The circle and the line intersect at $x$ given by $x^2+(x-k)^2=4$, i.e. $2x^2-2kx+k^2-4=0$.
There is only one possible $x$ if $\Delta=32-4k^2=0$, that is when $k=\pm2\sqrt{2}$.
Then the coordinates of the tangent points are given by $x=\frac k2, y=-\frac k2-2$.