Example 5.1.1. Tangent to a circle.
Solution 1. Solution A
To find the equation of a line we need a point and a slope. We already have the point at To find the slope, we can express the circle as the graph of 2 functions. We first solve for
We then take the square root to produce 2 functions.
The point is on the first function, which is the top half of the circle, so we take its derivative and evaluate at
Thus the tangent line, in point-slope form, is:
Solution 2. Solution B
To find the equation of a line we need a point and a slope. We already have the point at To find the slope, we take the derivative of our equation. Since we do not have y as a function of we simply note that its derivative is the placeholder Recall that the derivative of with respect to is simply 1.
We then solve for and substitute our point in for
When we substitute our point in for we get the same value, Thus the tangent line, in point-slope form, is: