# Circles Collision Detection

This topic is 4159 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

Hi guys. I just signed up here and I have a question that I wanted to ask. I’d really appreciate it if somebody could please help me with this. Imagine a circle with the radius of 2 inches. Now imagine another circle with the same radius drawn somehow that the center of the second circle is a point on the circumference of the first circle. Therefore, the center of Circle1 will be on the circumference of Circle2, also. Now what I need to know is the X and Y coordinates of the two points that are the collision points of the circumference of both of the circles. When these two circles collide, they do in two points that is obviously located on the circumference of both of them. Now is there any way that I can find the X and Y coordinates of these two points? I have tried the Pythagorean Theorem; finding the sine and cosine of some of the angles; length of the lines; midpoints; slope-intercept formulas and etc for each of the lines but no dice yet in finding the coordinates of the colliding points. I have attached a picture to this post that shows exactly what I need to find out: which is the coordinates of the “z” point in Cartesian coordinate system. I’d appreciate it if somebody could help me. Radius of the circles in the picture = 2.23607 http://www.asmtrauma.com/Circles.png

##### Share on other sites
Use the equation for a circle, (x-a)^2 + (y-b)^2 = r^2, where (a,b) is the center and r is the radius. Take the equations for the two circles and solve for x and y. You'll end up with a quadratic equation whose solutions will be the intersection points.

##### Share on other sites
If you only care about the collisions along the circumference, then take the equation for the circle for each and solve it together:

(x-h)^2 + (y-k)^2 - r^2 = 0

[(x-h)^2 + (y-k)^2 - r^2]1 = [(x-h)^2 + (y-k)^2 - r^2]2

##### Share on other sites
Thanks guys. I solved the system of linear equations and I got these results:

Circle1.Center = (0, 0), Radius = 2
Circle2.Center = (2, 0), Radius = 2

Circle1.Equation = x^2 + y^2 = 4
Circle2.Equation = (x-2)^2 + y^2 = r^2 = x^2 - 4x - 4 + y^2 = 4

Multiply the equation of the first circle by (-1):

-x^2 - y^2 = -4
x^2 - 4x - 4 + y^2 = 4

-4x - 4 = 0
-4x = 4
x = -1

Solve for y:

x^2 + y^2 = 4
1 + y^2 = 4
y^2 = 3
y = Sqrt(3)
y = 1.732050807568877293573725295594556428114

So the first collision point on the circumference of both of the circles would be (1, 1.73). Now what I need to know is the second point of collision. How can I find that? I'd appreciate your help.

##### Share on other sites
Quote:
 Original post by ExchangeSolve for y:x^2 + y^2 = 41 + y^2 = 4y^2 = 3y = Sqrt(3)y = 1.732050807568877293573725295594556428114So the first collision point on the circumference of both of the circles would be (1, 1.73). Now what I need to know is the second point of collision. How can I find that? I'd appreciate your help.

Remember that y^2 = 3 has two solutions: +sqrt(3) and -sqrt(3).

##### Share on other sites
Oh thanks. I don't know what I was thinking. Appreciations.

1. 1
2. 2
Rutin
16
3. 3
4. 4
5. 5

• 13
• 26
• 10
• 11
• 9
• ### Forum Statistics

• Total Topics
633722
• Total Posts
3013546
×