Posted by **Student** on Tuesday, October 16, 2007 at 6:29pm.

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??

- Ap Calc AB -
**Reiny**, Tuesday, October 16, 2007 at 6:47pm
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

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