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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??

1 answer

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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