solve, find the equations of the tangent line to the circle $$ x^2+y^2-8x+10y-128=0 $$ at the points of the circle on the x-axis.

To find the equations of the tangent lines to the circle at the points where the circle intersects the x-axis, we'll follow these steps:

### Step 1: Identify the equation of the circle
The given equation of the circle is:

\[
x^2 + y^2 - 8x + 10y - 128 = 0
\]

We can rewrite this equation in the standard form of a circle by completing the square.

### Step 2: Complete the square
First, we'll group the $x$ terms and the $y$ terms:

\[
(x^2 - 8x) + (y^2 + 10y) = 128
\]

Now, complete the square for both $x$ and $y$:

\[
(x^2 - 8x + 16) + (y^2 + 10y + 25) = 128 + 16 + 25
\]

This simplifies to:

\[
(x - 4)^2 + (y + 5)^2 = 169
\]

The equation represents a circle with:
- Center $(h, k) = (4, -5)$
- Radius $r = \sqrt{169} = 13$

### Step 3: Find the points where the circle intersects the x-axis
The points on the x-axis have $y = 0$. Substitute $y = 0$ into the circle's equation:

\[
(x - 4)^2 + (0 + 5)^2 = 169
\]

This simplifies to:

\[
(x - 4)^2 + 25 = 169
\]

Subtract 25 from both sides:

\[
(x - 4)^2 = 144
\]

Taking the square root of both sides:

\[
x - 4 = \pm 12
\]

So:

\[
x = 16 \quad \text{or} \quad x = -8
\]

The points where the circle intersects the x-axis are $(16, 0)$ and $(-8, 0)$.

### Step 4: Find the slopes of the tangent lines
The general formula for the slope of the tangent line to a circle at a point $(x_1, y_1)$ is given by:

\[
\text{slope} = -\frac{x_1 - h}{y_1 - k}
\]

#### For the point $(16, 0)$:
\[
\text{slope} = -\frac{16 - 4}{0 - (-5)} = -\frac{12}{5}
\]

#### For the point $(-8, 0)$:
\[
\text{slope} = -\frac{-8 - 4}{0 - (-5)} = \frac{12}{5}
\]

### Step 5: Write the equations of the tangent lines
The equation of a line with slope $m$ passing through a point $(x_1, y_1)$ is:

\[
y - y_1 = m(x - x_1)
\]

#### For the point $(16, 0)$ and slope $-\frac{12}{5}$:
\[
y - 0 = -\frac{12}{5}(x - 16)
\]
\[
y = -\frac{12}{5}x + \frac{192}{5}
\]

#### For the point $(-8, 0)$ and slope $\frac{12}{5}$:
\[
y - 0 = \frac{12}{5}(x + 8)
\]
\[
y = \frac{12}{5}x + \frac{96}{5}
\]

### Final Answer:
The equations of the tangent lines to the circle at the points on the x-axis are:

\[
y = -\frac{12}{5}x + \frac{192}{5}
\]

\[
y = \frac{12}{5}x + \frac{96}{5}
\]

Would you like any further details or explanations?

---

Here are 5 related questions for further exploration:

1. How do you find the equation of a tangent line to a circle at a given point not on the x-axis?
2. What is the significance of the perpendicular distance from the center to the tangent line?
3. Can you find the points of tangency on the y-axis for the same circle?
4. How would you derive the general formula for the slope of a tangent to a circle?
5. What happens to the tangent lines if the circle is shifted vertically?

**Tip:** Remember, the tangent to a circle is always perpendicular to the radius at the point of tangency.

Learn how to find equations of tangent lines to a circle at points where it intersects the x-axis. Includes step-by-step solution using circle equations and completing the square method. Suitable for Grades 10-12 students studying circle geometry and algebra.

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