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

## Answer This Question

## Related Questions

- secant line, tangent line - f(x) = sqrt(x-1), 1<=x<=3 Let A = (a,f(a)), ...
- calc - Find an equation of the tangent line to the curve at the given point. y...
- Calc. - sketch the curve using the parametric equation to plot the points. use ...
- Calculus AB - Could someone please help me with these tangent line problems? 1) ...
- Math - Find the equation of the line tangent to curve x=sec(t), y=tan(t), at t=...
- Calculus - Please help this is due tomorrow and I don't know how to Ive missed a...
- Calculus - If F(x)=x^3−7x+5, use the limit definition of the derivative to...
- Calculus - Tangent Line - Hi, im having problems with the following problem. The...
- Help Calc.! - original curve: 2y^3+6(x^2)y-12x^2+6y=1 dy/dx=(4x-2xy)/(x^2+y^2+1...
- please help; calc - original curve: 2y^3+6(x^2)y-12x^2+6y=1 dy/dx=(4x-2xy)/(x^2+...

More Related Questions