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To find the equation of the tangent line to the circle given by the equation \( x^2 + y^2 - 4x + 2y - 8 = 0 \) at the point \( P(0, 2) \), we first need to confirm that the point \( P(0, 2) \) lies on the circle.

1. **Verify the Point on the Circle:**

Substitute \( (0, 2) \) into the equation of the circle:

\[ (0)^2 + (2)^2 - 4(0) + 2(2) - 8 = 0 + 4 + 0 + 4 - 8 = 0 \]

Since the equation holds true, the point \( P(0, 2) \) is indeed on the circle.

2. **Find the Gradient (Derivative) at the Point:**

Rewrite the equation of the circle in a more familiar form. However, for simplicity, we'll directly use implicit differentiation to find the slope of the tangent line.

Starting with the equation of the circle:

\[ x^2 + y^2 - 4x + 2y - 8 = 0 \]

Differentiate both sides with respect to \( x \):

\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) - \frac{d}{dx}(4x) + \frac{d}{dx}(2y) - \frac{d}{dx}(8) = 0 \]

\[ 2x + 2y \frac{dy}{dx} - 4 + 2 \frac{dy}{dx} = 0 \]

Combine the terms with \( \frac{dy}{dx} \):

\[ 2y \frac{dy}{dx} + 2 \frac{dy}{dx} = 4 - 2x \]

\[ (2y + 2) \frac{dy}{dx} = 4 - 2x \]

\[ \frac{dy}{dx} = \frac{4 - 2x}{2(y + 1)} \]

Substitute \( (x, y) = (0, 2) \):

\[ \frac{dy}{dx} \bigg|_{(0, 2)} = \frac{4 - 2(0)}{2(2 + 1)} = \frac{4}{6} = \frac{2}{3} \]

The slope of the tangent line at \( P(0, 2) \) is \( \frac{2}{3} \).

3. **Equation of the Tangent Line:**

Using the point-slope form of the equation of a line \( y - y_1 = m(x - x_1) \):

\[ y - 2 = \frac{2}{3}(x - 0) \]

\[ y - 2 = \frac{2}{3}x \]

\[ y = \frac{2}{3}x + 2 \]

Therefore, the equation of the tangent line at point \( P(0, 2) \) is:

\[ y = \frac{2}{3}x + 2 \]

Would you like further details on this solution or have any additional questions?

Here are 5 more questions you might find interesting:
1. How do you find the center and radius of a circle from its equation?
2. Can you explain how to derive the equation of a tangent line to a curve?
3. What is implicit differentiation and when is it used?
4. How can you determine if a point lies on a given curve or circle?
5. What is the geometric interpretation of the derivative at a point on a curve?

**Tip:** When dealing with circles and tangents, verifying that a point lies on the circle before proceeding with differentiation helps ensure accuracy in your calculations.

Learn how to find the equation of the tangent line to a circle at a given point. This problem involves circle geometry, implicit differentiation, and point-slope form of a line. Suitable for advanced high school students.

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