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.