What is the equation of the line tangent to the circle of x^2+y^2=8 at the point (2, 2)?

Answer 1

We want to find the equation of a line y = mx + b; and we know the point (2,2) is on that line. Recall, the derivative y' is the slope (m) of the tangent line at that point (2,2). What this means is for our circle x² + y² = 8, we must use implicit differentiation to find y' and that will give us m:

x² + y² = 8

Differentiate both sides. 2x + 2y·y' = 0

Substitute x =2 and y = 2 2(2) + 2(2)·y' = 0

4 + 4·y' = 0

Solve for y' y' = -1

Now that we know our slope m = -1, the tangent line is y = (-1)x + b for some value b. To figure out what be is, we again substitute x = 2 and y = 2:

2 = (-1)(2) + b

Solve for b b = 4

Therefore, the tangent line at (2,2) is given by the equation y = -x + 4