point deflecting off of a circle

I know that this is a very simple question, but I am stuck at the very last part of this process and can't find the solution elsewhere (I figured I'd find it on this site, but I didn't see it).

I have an object that is colliding with a circle and I need it to deflect properly, like this:
I know the coordinates of the center of the circle and the object when it is on the circle's perimeter. I know the direction that the object is traveling on contact and can calculate the direction to the center (pointing inwards).

From similar questions, I know that the tangent line is perpendicular to the radius line I calculate. But, I'm not sure where to go after that. I need to calculate the new direction of the object in degrees, but my idea $\theta = \theta + 2(radiusline - \theta)$ , where $radiusline$ is the vector pointing towards the center, is inaccurate.

What is the proper formula for this deflection?

• use the tangent to calculate the normal at the point of contact, and the angle the direction vector makes with it. The reflection vector is then simple to calculate. – K. Rmth Dec 8 '13 at 21:07
• I have the direction vector. Is the tangent vector (since it points in a direction) equal to the direction plus 90 or minus 90? The normal at the point of contact is just the line to the radius, isn't it? – person27 Dec 8 '13 at 21:33
• Aren't mixing two things up? The direction vector applies to the moving object whilst the tangent vector applies to the circle. Using the normal implies the benefit of simplicity; the normal vector(direction towards the circle centre) at a point of incidence $(b,c)$ is just $\vec N=-b\underline{\hat i}-c\underline{\hat j}$ – K. Rmth Dec 8 '13 at 21:52
• I don't understand what the carot over the i is, what the i and j stand for, or what the underscores are. I have the math education and capacity of your average high-school student. You could say I'm definitely not mathematically-minded. – person27 Dec 8 '13 at 21:52
• I will write an answer to this. Hopefully it will no longer be greek to you then.$\hat i$ is the unit vector in the positive x-direction (horizontal) and $\hat j$ is likewise for the y-direction (vertical). – K. Rmth Dec 8 '13 at 21:53

\begin{align} \vec N & = \text{normal at point of incidence}=-a \hat{\mathbf{i}}-b \hat{\mathbf{j}} \\ \vec V & = \text{incident vector}=u \hat{\mathbf{i}}+v \hat{\mathbf{j}} \\ \vec R & = \text{reflected vector}=c \hat{\mathbf{i}}+d \hat{\mathbf{j}} \\ \end{align} \begin{align} \text{using}\ \vec R =\vec V -2\vec N(\vec V \cdot \vec N)&={u \choose v}-2{-a \choose -b}\left[{u \choose v}\cdot {-a \choose -b}\right]\\ &\\ &={u \choose v}-2(au-bv){-a \choose -b}\\ &={u+2a^2u-2abv \choose v+2a^2u-2b^2v} \equiv {c \choose d}\\ \end{align} Hence $\vec R=(u+2a^2u-2abv)\hat{\mathbf{i}}+(v+2a^2u-2b^2v)\hat{\mathbf{j}}$