To solve the given vector line integral, we proceed as follows:

Given Problem:

We are to compute the vector line integral:

$\int_C \left( \frac{-y}{(x-5)^2 + y^2} \, dx + \frac{x-5}{(x-5)^2 + y^2} \, dy \right)$

where $C$ is the circle of radius 3 centered at $(5, 0)$, oriented counterclockwise.

Step 1: Recognize the Vector Field

The vector field is:

$\mathbf{F}(x, y) = \left( \frac{-y}{(x-5)^2 + y^2}, \frac{x-5}{(x-5)^2 + y^2} \right).$

Step 2: Parameterize the Circle

The circle $C$ of radius 3 centered at $(5, 0)$ can be parameterized as:

$x(t) = 5 + 3\cos t, \quad y(t) = 3\sin t, \quad t \in [0, 2\pi].$

From this parameterization: $dx = -3\sin t \, dt, \quad dy = 3\cos t \, dt.$

Step 3: Substitute into the Integral

Substitute $x(t)$, $y(t)$, $dx$, and $dy$ into the given line integral:

$\int_C \left( \frac{-y}{(x-5)^2 + y^2} \, dx + \frac{x-5}{(x-5)^2 + y^2} \, dy \right).$

1. Compute $(x-5)^2 + y^2$: $(x-5)^2 + y^2 = (5 + 3\cos t - 5)^2 + (3\sin t)^2 = (3\cos t)^2 + (3\sin t)^2 = 9(\cos^2 t + \sin^2 t) = 9.$

2. Rewrite the terms of the vector field: $\frac{-y}{(x-5)^2 + y^2} = \frac{-3\sin t}{9} = -\frac{\sin t}{3}, \quad \frac{x-5}{(x-5)^2 + y^2} = \frac{3\cos t}{9} = \frac{\cos t}{3}.$

3. Substitute $dx$ and $dy$: $dx = -3\sin t \, dt, \quad dy = 3\cos t \, dt.$

Now the integral becomes: $\int_0^{2\pi} \left( \left(-\frac{\sin t}{3}\right)(-3\sin t \, dt) + \left(\frac{\cos t}{3}\right)(3\cos t \, dt) \right).$

Step 4: Simplify the Integral

Simplify each term:

1. First term: $\left(-\frac{\sin t}{3}\right)(-3\sin t) = \sin^2 t.$

2. Second term: $\left(\frac{\cos t}{3}\right)(3\cos t) = \cos^2 t.$

Thus, the integral becomes: $\int_0^{2\pi} \left( \sin^2 t + \cos^2 t \right) dt.$

Step 5: Use the Pythagorean Identity

Using $\sin^2 t + \cos^2 t = 1$, the integral simplifies to: $\int_0^{2\pi} 1 \, dt = \left[ t \right]_0^{2\pi} = 2\pi.$

Final Answer:

The value of the line integral is: $\boxed{2\pi}.$

Let me know if you have questions or need clarification!

Here are 5 related questions for further exploration:

1. How is the parameterization of a circle derived for different centers?
2. How does Green's theorem apply to vector line integrals like this one?
3. What happens if the circle's orientation is reversed to clockwise?
4. Can this integral be computed using polar coordinates directly?
5. How does the result change if the radius of the circle is altered?

Tip: Always simplify the vector field before substituting the parameterization—it often makes the calculation easier!