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Let's start by analyzing and solving the problem.

The given circle is:
\[
x^2 + y^2 = 25
\]
which is a circle centered at the origin \((0, 0)\) with radius 5, since \(\sqrt{25} = 5\).

### Part (a): Find the equation of the tangent line to the circle at the point \((3, -4)\).
The slope of the radius at the point \((3, -4)\) is the slope from the origin to this point, which is:
\[
\text{slope of radius} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-4 - 0}{3 - 0} = \frac{-4}{3}
\]
Since the tangent line is perpendicular to the radius, the slope of the tangent line is the negative reciprocal of \(-\frac{4}{3}\), which is:
\[
\text{slope of tangent line} = \frac{3}{4}
\]
Using the point-slope form of a line equation, \(y - y_1 = m(x - x_1)\), with slope \(\frac{3}{4}\) and point \((3, -4)\), we get:
\[
y - (-4) = \frac{3}{4}(x - 3)
\]
Simplifying:
\[
y + 4 = \frac{3}{4}(x - 3)
\]
\[
y = \frac{3}{4}x - \frac{9}{4} - 4
\]
\[
y = \frac{3}{4}x - \frac{25}{4}
\]
Thus, the equation of the tangent line is:
\[
y = \frac{3}{4}x - \frac{25}{4}
\]

### Part (b): Find the other point on the circle where a tangent line is parallel to the tangent line in part (a).
For the tangent line to be parallel, it must have the same slope, \(\frac{3}{4}\).

We know the equation of the circle is \(x^2 + y^2 = 25\). The slope of the tangent line at any point on the circle is the negative reciprocal of the slope of the radius at that point. The slope of the radius at any point \((x, y)\) is \(\frac{y}{x}\), so the slope of the tangent line is \(-\frac{x}{y}\).

To find the point where the slope of the tangent line is \(\frac{3}{4}\), we set \(-\frac{x}{y} = \frac{3}{4}\), which gives:
\[
-\frac{x}{y} = \frac{3}{4}
\]
\[
x = -\frac{3}{4}y
\]
Now, substitute this into the equation of the circle \(x^2 + y^2 = 25\):
\[
\left(-\frac{3}{4}y\right)^2 + y^2 = 25
\]
\[
\frac{9}{16}y^2 + y^2 = 25
\]
\[
\frac{9}{16}y^2 + \frac{16}{16}y^2 = 25
\]
\[
\frac{25}{16}y^2 = 25
\]
\[
y^2 = 16
\]
\[
y = \pm 4
\]
If \(y = 4\), then \(x = -\frac{3}{4}(4) = -3\). So one point is \((-3, 4)\).

Thus, the other point where the tangent line is parallel to the one found in part (a) is \((-3, 4)\).

---

Would you like any further details or clarifications? Here are some related questions:

1. How can we find the equation of a tangent line to any circle?
2. Why is the slope of the tangent line the negative reciprocal of the slope of the radius?
3. How does the geometry of a circle influence the relationship between tangent lines and radii?
4. What are other methods to find the points of tangency on a circle?
5. How would the solution change if the center of the circle was not at the origin?

**Tip**: For any circle, the slope of the tangent line at a point can be derived directly from the negative reciprocal of the radius's slope at that point.

Find an equation for the line tangent to the circle x^2 + y^2 = 25 at the point (3, -4). At what other point on the circle will a tangent line be parallel to the tangent line at (3, -4)?

This problem involves finding the equation of a tangent line to the circle x^2 + y^2 = 25 at the point (3, -4) and determining another point on the circle where the tangent line is parallel. It explores key geometric principles including the relationship between tangents, radii, and slopes.

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