# Solving for Orthogonal Trajectories

1. Sep 15, 2009

### vorse

1. The problem statement, all variables and given/known data

Which of the following is the set of orthogonal trajectories for the family indicated by

(x-c)^2 + y^2 = c^2

a). (x-c)^2 + y^2 = c^2

b). (x-c)^2 - y^2 = c^2

c). x^2 + (y-c)^2 = c^2

d). x^2 - (y-c)^2 = c^2

e). None of the above

2. Relevant equations

I think this is a homogenous equation after playing with the variables and substituting for C. I came up with:

y' = (y^2 - x^2)/xy

I then substitute y=xv after setting y' = 1/y' for the orthogonal trajectories and tried to solve the equation, but ended up with answers no where close to the choices.

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 15, 2009

### tiny-tim

Hi vorse!

Let's do it the easy way

draw a few curves, and that should show you the most obvious orthogonal trajectory …

which of a) to e) does that one lie on?

3. Sep 15, 2009

### vorse

I was afraid you'd said I have to graph the equation; I can never make heads to tail of what I draw out.

4. Sep 16, 2009

### tiny-tim

ok … (x-c)2 + y2 = c2 is obviously a circle.

So, for each value of c, what is its centre, what is its radius, and what are the endpoints of its "horizontal" diameter?

5. Sep 16, 2009

### vorse

Ok, I figured out the answer. I think it's answer choice C. It's just my answer has something like 2c as one of the constant, and I guess the answer choice reduce it down to just C. Btw, I didn't graph the equation; it is a homogenous equation and can be solve that way; just the algebra manipulation was confusing. Thanks for all the help PF. Thanks for trying to help me tiny tim.

Last edited: Sep 16, 2009
6. Sep 17, 2009

### tiny-tim

Yup!

But a quick sketch always helps …

in this case, the original curves are the circles with centres on the x-axis, all touching each other (and the y-axis) at the origin.

Quickly hovering over this with a pencil shows that the orthogonal trajectories must approach the y-axis horizontally, and must all go through the origin …

which choice C does, since it's the circles with centres on the y-axis.

hmm … are those all of the orthogonal trajectories?

(btw, the wiki page http://en.wikipedia.org/wiki/Orthogonal_trajectories" [Broken])

Last edited by a moderator: May 4, 2017