# Find the radius of the osculating circle

## Homework Statement

Let C be the curve of intersection of the cylinder x^2-2x+y^2 =3 and the plane z=y+1. Find the radius of the osculating circle of the curve C at point (1,2,3)

## The Attempt at a Solution

I am not really sure how to start it...
i tried finding the point of intersection by setting y=0 and got z=1, x=3, but doesnt really make sense to me to do so. I also tried setting the equation to each other, but it doesnt seem to make sense too..

it would be appreciated if anyone can tell me how to begin and guide me through it

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Dick
Homework Helper
Your first job is to find a parametric expression for the curve r(t)=(x(t),y(t),z(t)). Write the first equation as (x-1)^2+y^2=2. That's the equation of a circle with center (1,0) and radius sqrt(2), right? Can you write expressions for x and y in terms of cos(t) and sin(t) where t is the angle the point makes with the circle center? Now can you use z=y+1 to write z in terms of t? Now I'm guessing you probably know how to find the radius from the parametric expression using the curvature.

Your first job is to find a parametric expression for the curve r(t)=(x(t),y(t),z(t)). Write the first equation as (x-1)^2+y^2=2. That's the equation of a circle with center (1,0) and radius sqrt(2), right? Can you write expressions for x and y in terms of cos(t) and sin(t) where t is the angle the point makes with the circle center? Now can you use z=y+1 to write z in terms of t? Now I'm guessing you probably know how to find the radius from the parametric expression using the curvature.
I think i get it, but here is what i got

re parametrize the intersection, and i got c=(x=2cost+1,y=2sint, z= 2sint+1) and point (1,2,3) is t=pi/2
and to find radius is R=1/curvature at t=pi/2?

Dick