We are given a circle with its center at $(3, -5)$ and a radius of $7$. The general equation of a circle with center $(h, k)$ and radius $r$ is:

$(x - h)^2 + (y - k)^2 = r^2$

Substituting the given values $h = 3$, $k = -5$, and $r = 7$, the equation becomes:

$(x - 3)^2 + (y + 5)^2 = 7^2$ $(x - 3)^2 + (y + 5)^2 = 49$

We are interested in the points where the circle crosses the x-axis. At these points, the y-coordinate is 0, so we substitute $y = 0$ into the equation:

$(x - 3)^2 + (0 + 5)^2 = 49$ $(x - 3)^2 + 25 = 49$ $(x - 3)^2 = 24$

Now, solve for $x$ by taking the square root of both sides:

$x - 3 = \pm \sqrt{24}$ $x - 3 = \pm 2\sqrt{6}$ $x = 3 \pm 2\sqrt{6}$

Thus, the two x-coordinates where the circle intersects the x-axis are:

$x_1 = 3 + 2\sqrt{6} \quad \text{and} \quad x_2 = 3 - 2\sqrt{6}$

The distance $pq$ between the points $P$ and $Q$ on the x-axis is the absolute difference between these two x-coordinates:

$pq = |(3 + 2\sqrt{6}) - (3 - 2\sqrt{6})|$ $pq = |3 + 2\sqrt{6} - 3 + 2\sqrt{6}|$ $pq = |4\sqrt{6}|$ $pq = 4\sqrt{6}$

So, the distance $pq$ is $4\sqrt{6}$ units.

Would you like further details or any clarification on this solution?

Here are some related questions for further exploration:

1. How would the equation of the circle change if the center were at $(0, 0)$?
2. How can you find the intersection points of a circle with the y-axis?
3. What would happen if the radius of the circle was smaller than the distance from the center to the x-axis?
4. Can you derive the equation of the circle if you only know two points on the circle and its radius?
5. How do you calculate the distance between two points on a circle, given the coordinates of the points?

Tip: Always substitute $y = 0$ to find intersections with the x-axis, and $x = 0$ for intersections with the y-axis.