Thread: Curve's point A tangent line passes through a point P. What is the distance of AP

Let A be a point on the curve C: . If the tangent line at A passes through P(4,3), then the length of AP is?

So i'm stuck midway at the solution for this one.
So far I've gotten these equations
; the tangent line slope at any point on the curve (by implicit differentiation)
; the slope at point P; acquired by y-k=m(x-h)

and when I put those two equations together I get:


And when I input this equation back into the original curve i end up with
x= -y

I'm stuck at this point because when I plug in the last equation back to original curve again, I'm getting an complex solution.
Help please!

Oh Gosh. I am sorry. That's supposed to be

You sure you got the curve C right. Completing the square in gives



There are no real solutions to this equation.

Originally Posted by Danny

You sure you got the curve C right. Completing the square in gives



There are no real solutions to this equation.

Oh Gosh. I am sorry. That's supposed to be
{x}^{ 2} + {y}^{ 2} - 2x - 4 = 0

Originally Posted by gundanium

Let A be a point on the curve C: . If the tangent line at A passes through P(4,3), then the length of AP is?

...

1. Your curve is a circle with the center M(1, 0) and the radius .

2. The triangle MPA is a right triangle with MP as the hypotenuse and MA = r.

3. Use Pythagorean theorem to determine the length of AP.

Spoiler:

This is the way I would do it. Suppose the point on the circle is Then satisfies

.

The tangent to the curve at this point is



and the slope of the line connecting to is



These must equal, hence



or



You are required to find the distance between the two points or




Now gives



so



Edit: Earthboth's solution is so much more elegant (and of course easier!)

Originally Posted by Danny

Now gives

Why? How did you come up with this?

This is what I have





this is what I want



Multiply (LHS only) and equate like terms. As it turns out, there is a unique and solution.