Find the equation of the line tangent to the curve at (5, -3)
(x-2)^2 + (y+3)^2 = 9
I solved the derivative to be dy/dx = ((-2x+4)/ (2y+6))
when i plugged in the points (5, -3) I got the slope as -6/0...How is this possible??
How can i find the equation of this curve if the slope is undefined??
The original equation is the equation of a circle with centre at (2,-3) and a radius of 3
So a tangent at (5,-3) would be a vertical line. Make a rough sketch to see what I mean.
The equation of any vertical line is x = c
in your case
x = 5