Question

Find the equation of the tangent line for the curve x^2 + y^2 = 4 at the point (2, 0)

Answer 1

To find the equation of the tangent line to the curve given by the equation \(x^2 + y^2 = 4\) at the point \((2, 0)\), we will first need to calculate the derivative of the function implicitly and then evaluate it at the point of tangency.

1. **Differentiate the equation implicitly:**

Start with the equation of the circle:
\[
x^2 + y^2 = 4
\]

Differentiate both sides with respect to \(x\):
\[
\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(4)
\]
This gives:
\[
2x + 2y \frac{dy}{dx} = 0
\]

Now, solve for \(\frac{dy}{dx}\):
\[
2y \frac{dy}{dx} = -2x \implies \frac{dy}{dx} = -\frac{x}{y}
\]

2. **Evaluate the derivative at the point (2, 0):**

Substitute \(x = 2\) and \(y = 0\) into the derivative:
\[
\frac{dy}{dx} = -\frac{2}{0}
\]
Here, we have division by zero, which means that the slope of the tangent line at this point is undefined. This indicates that the tangent line is vertical.

3. **Find the equation of the tangent line:**

Since the slope is undefined, the equation of the vertical line passing through the point \((2, 0)\) is:
\[
x = 2
\]

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

\[
\boxed{x = 2}
\]