Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have to find the necessary translations in X and Y to move a point 0n a circle to another one.

I have a center (X and Y coordinates), a radius, and a current position in radians. And given a value in radians (the amount I want to translate), I have to find the amount of values in X and Y I have to move to get to that position.

So, for example, if default values are (2, 3) for the center, the radius is 3 and the starting radian position is 0. The starting point will be (5, 3) and I if I want to move the position 0.2 radians, I want to know how many units of X and Y I have to make to go to that position.

This is a simple drawing of what I want to do. Also, check out which one is the 0 radian starting position (rightmost part)

enter image description here

share|improve this question
add comment

3 Answers

up vote 0 down vote accepted

Let us first work with the example you gave and generalize from there.

You have a circle with center $(2,3)$ and radius $r=3$. You want to rotate the point $(5,3)$ on the circle by $\theta=0.2$ radians. To do this we parametrize the circle as $(x,y)=(2+3\cos\theta,3+3\sin\theta)$. The point $(5,3)$ has $\theta=0$ and we want to increase that angle by $0.2$. Thus the new point is $(2+3\cos(0.2),3+3\sin(0.2))\approx(4.9,3.6)$.

Now, in the general case, say you have a circle with center $(a,b)$ and radius $r$. The position of the initial point is $\theta$ radians along the circle from $(a+r,b)$. The parametric equation for the circle is $(x,y)=(a+r\cos\theta,b+r\sin\theta)$. Say you want to increase by $\phi$ radians. Then the new point is $$(a+r\cos(\theta+\phi),b+r\sin(\theta+\phi))$$

share|improve this answer
add comment

Using center $(2, 3) = (x_c, y_c)$ and knowing radius = $r$,

  • you can compute both the initial position of the point at $\theta = 0$ radians ($\theta$ = the angle formed by the radius, with respect to the positive x-axis),
  • and the location of the translated point when $\theta$ is rotated counter-clockwise by $0.2$ radians.

$$x - x_c = r\cos \theta:\quad x - 2 = 3 \cos \theta$$

$$y - y_c = r \sin \theta:\quad y - 3 = 3 \sin \theta$$

Substitute the value of the rotation angle into theta to evaluate for the desired $(x, y)$

At $\theta = 0$, $\;x = 2 + 3 = 5$, $\;y = 3 + 0 = 3.$ Starting point = $(5, 3)$

At $\theta = 0.2,\;$ $x = 2 + 3\cos(0.2)$, $\;y = 3 + 3\sin(0.2)$.

Ending point = $(2 + 3\cos(0.2), 3 + 3\sin(0.2)) \approx (4.9402, 3.5960).$

In general, if you have a circle with center $(x_0,y_0)$ and radius $r$, the parametric equation for the circle is $(x,y)=(x_0+r\cos\theta,y_0+r\sin\theta)$. So at $\theta = 0$, that gives us the point $(x_0 + r, y_0)$. If you want to find a point on the circle at $\theta > 0$ radians, the new point is $$(x_0+r\cos(\theta),y_0+r\sin(\theta)).$$


To see why we can write $x, y$ in terms of $\;\theta\;$ and $\;r\;$ (image assumes center = $(0,0))$:

enter image description here

The point where the end of the radius meets the circle is given by $(x = 0 + r\cos\theta, y = 0 + r\sin\theta) = (x = r\cos\theta, y = r\sin\theta)$.

share|improve this answer
add comment

I'm interpreting "translation by $\theta$ radians" to mean a rotation by $\theta$ radians acticlockwise.

We know that a rotation of $(x, y)$ about the origin by $\theta$ radians anti clockwise, brings it to the point $ ( x \cos \theta - y \sin \theta, x \sin \theta + y \cos \theta)$, which you can verify manually.

Hence, to answer your questions, we first translate the center of the circle to the origin, apply the rotation ("translation") and then translate the origin back to the center. So, if the point is $(x,y)$ and the center of the circle is $(a,b)$, the first translation brings us to the points $(x-a, t-b)$, the next rotation brings us to the point $( (x-a) \cos \theta - (y-b) \sin \theta, (x-a) \sin \theta + (y-b) \cos \theta)$ and the last translation bring it to the point

$$( (x-a) \cos \theta - (y-b) \sin \theta +a , (x-a) \sin \theta + (y-b) \cos \theta +b)$$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.