[SOLVED] Setting a slope equal to a circle's derivative

kedinik

I'm working on a personal programming project and I've hit a roadblock with a collision algorithm.

I'm trying to find the 2 points on a circle where the slope of the circle equals a given constant.

How do I solve m = -(x/y) for the two x values that correspond to the solutions? When I try to substitute in (r^2 - x^2)^(1/2) for y, the best I can manage is a messy quartic.

Seol

kedinik wrote: »

I'm working on a personal programming project and I've hit a roadblock with a collision algorithm.

I'm trying to find the 2 points on a circle where the slope of the circle equals a given constant.

How do I solve m = -(x/y) for the two x values that correspond to the solutions? When I try to substitute in (r^2 - x^2)^(1/2) for y, the best I can manage is a messy quartic.

The tangent of a circle's curve is perpendicular to the radius. So, take the normal of the slope you want, and the points you want will be a vector equal to travelling along the normal by the circle's radius from its centre.

kedinik

Seol wrote: »

kedinik wrote: »

I'm working on a personal programming project and I've hit a roadblock with a collision algorithm.

I'm trying to find the 2 points on a circle where the slope of the circle equals a given constant.

How do I solve m = -(x/y) for the two x values that correspond to the solutions? When I try to substitute in (r^2 - x^2)^(1/2) for y, the best I can manage is a messy quartic.

The tangent of a circle's curve is perpendicular to the radius. So, take the normal of the slope you want, and the points you want will be a vector equal to travelling along the normal by the circle's radius from its centre.

That's really clever! I think I may have just figured out the formulaic solution, but your way is much cleaner.

Thank you!

enlightenedbum

Take the equation of your circle:

(x-a)^2 + (y-b)^2 = r^2

Differentiate everything with respect to x:

2 * (x -a) + 2 * (y - b) * (dy/dx) = 0

Solve for dy/dx

2 * (x - a) = -2 * (y-b) (dy/dx)

- (x - a) / (y - b) = dy/dx

So yeah, you're going to end up with a messy quartic.

- (x - a) / (r^2 - x^2) ^ (1/2) = m

Fractions are evil, so destroy them:

- (x - a) = m * (r^2 - x^2) ^ (1/2)

Radicals are also evil...

(x - a) ^ 2 = m ^ 2 (r^2 - x ^2)

Expand:

x ^2 - 2 * a * x + a^2 = m^2 * r^2 - m^2 * x ^ 2

(m^2 + 1) * x ^ 2 - 2 * a * x + a^2 - m^2 * r^2 = 0

Apply quadratic formula, ta da!

kedinik

Thank you too, enlightenedbum. Nice Colonization LP in G&T, btw, I've been enjoying it.

GoodOmens

I think it might be easier to do it this way:
Translate the circle so that its center is at the origin (so that you can ignore a and b).
Find the location on the new circle using x^2 + y^2 = r^2 and differentiating.
Then translate the circle back to its original location, which would involve translating that point as well.

That way, a and b don't end up as part of your calculation. More steps but a simpler computation than enlightendbum's (excellent and comprehensive) solution.