What is the rule for rotations about a point not the origin?

Question

I know the rules for $90^\circ$(counterclockwise and clockwise) rotations, and $180^\circ$ rotations, but those are only for rotations about the origin. What is the rule for a rotation above that is not about the origin?
By rule, I mean this: $(x, y) \rightarrow (y, -x)$.

Answer 1

I suppose there are lots of ways of looking at motions of the plane, but try this:

First, if you’re going to turn the plane about the origin through an angle of $\theta$ (positive for counterclockwise), then the rule is: $$ (x,y)\mapsto(x',y')=(x\cos\theta-y\sin\theta,x\sin\theta+y\cos\theta)\,. $$ That is, if your point $P=(x,y)$, the rotated point is $P'=(x',y')$.

Now if your center of rotation is not $(0,0)$ but rather $Q=(\alpha,\beta)$, you want to do the operation in three steps: first move your point $P=(x,y)$ leftward by $\alpha$ and downward by $\beta$. (You see that my mental picture is that $Q$ is in the first quadrant; but this is immaterial to the formulas.) This means the new point has coordinates $(x-\alpha,y-\beta)$. Now you apply your rotation of $\theta$, to get the rotated (after translation) point to have coordinates $\bigl((x-\alpha)\cos\theta-(y-\beta)\sin\theta,(x-\alpha)\sin\theta +(y-\beta)\cos\theta \bigr)$. Finally, you want to move everything back, but shifting rightward by $\alpha$ and upward by $\beta$. Therefore, your final formula is $$ (x,y)\mapsto (x',y')=\bigl(\alpha+(x-\alpha)\cos\theta-(y-\beta)\sin\theta, \beta+(x-\alpha)\sin\theta +(y-\beta)\cos\theta \bigr)\,. $$ For instance, if your center of rotation is $Q=(1,2)$ and you want to rotate $90^\circ$ counterclockwise about $Q$, the formula would be $$ (x,y)\mapsto\bigl(1-(y-2),2+(x-1)\bigr)=(3-y,1+x)\,. $$