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I am trying to calculate a point on a circle using an angle and a different point.

Circle

With this picture, I know the origin O, the radius r, the angle A, and the point B. Now I want to find the point C. How could I do this? (Note: the origin is not at (0,0))

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  • $\begingroup$ Hint : You can use the trigonométric identity about the sinus and the cosinus of a sum of two angles. $\endgroup$ – Jennifer May 28 '16 at 20:57
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I will use $\alpha$ instead of $A$. suppose first that O(0,0). Note that every point on the circle has the coordinates $(R\cos\theta,R\sin\theta)$. If $\theta_B$ is the angle of point B, then the angle $\theta_C$ of point C is $\theta_C=\theta_B-\alpha$. Hence, the coordinates of $C$ are $$ (R\cos(\theta_B-\alpha),R\sin(\theta_B-\alpha)) $$ that is $$ (R\cos\theta_B\cos\alpha+R\sin\theta_B\sin\alpha,R\sin\theta_B\cos\alpha-R\cos\theta_B\sin\alpha) $$ By setting $x_B=R\cos\theta_B$, $y_B=R\sin\theta_B$ we get $$ (x_B\cos\alpha+y_B\sin\alpha,y_B\cos\alpha-x_B\sin\alpha) $$ that is $$ \begin{pmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \\ \end{pmatrix} \begin{pmatrix} x_B \\ y_B \end{pmatrix} $$ in matrix form. Now, if $O\neq(0,0)$, say $O(o_1,o_2)$, then $$ \begin{pmatrix} o_1 \\ o_2 \end{pmatrix}+\begin{pmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \\ \end{pmatrix} \begin{pmatrix} x_B-o_1 \\ y_B-o_2 \end{pmatrix} $$ is the desired coordinates.

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  • $\begingroup$ What about the translation since the origin is not at $(0,0)$? $\endgroup$ – Carser May 28 '16 at 21:14
  • $\begingroup$ thanks Jed, i corrected it $\endgroup$ – boaz May 28 '16 at 21:25
  • $\begingroup$ Your answer gave me the solution I needed in a way that I could understand it. Thanks for including the origin shift. $\endgroup$ – Seth Welch May 29 '16 at 5:35
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There are several different approaches you can take. Since you're asking about trigonometry and polar coordinates, here's an approach:

  1. Find the angle to B from the horizontal.
  2. Find the angle to C by subtracting A.
  3. Find C using its angle with r

The angle to B can be found with $\theta_B = \sin^{-1}(B_y/B_x)$ where $B_y$ is the $y$ coordinate of B.

Then the angle to C is $\theta_C = \theta_B - A$

Finally, $C = (C_x,C_y) = (r \cos(\theta_C), r \sin(\theta_C)) $

If your origin $O$ is not at $(0,0)$, you can shift the whole thing down so that $O$ is temporarily at $(0,0)$, then you just shift your result back by the opposite amount.

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Don't use the word "origin"- use "center". Let the center of the circle be $(x_0, y_0)$. If point B is $(x_B, y_B)$ then the angle line OB makes with a horizontal line is $arcsin\left(\frac{y_B}{r}\right)$ and then the angle OC makes with the horizontal is $arcsin\left(\frac{y_B}{r}\right)- A$. That means that point C is $\left(r cos\left(arcsin\left(\frac{y_B}{r}\right)- A\right), r sin\left(arcsin\left(\frac{y_B}{r}\right)- A\right)\right)$

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If the center not being at $(0,0)$ is your problem, do an 'origin shift' by setting $$X = x-h \\ Y = y-k$$ where $(h,k)$ is the coordinate of the center.

Now you can work as if the center is at $(0,0)$ in your new coordinate system.


Parametric Form of Circle:

The parametric form of the equation of the circle is $$\frac{X-\alpha}{\cos \theta} = \frac{Y-\beta}{\sin \theta} = r $$

where $(\alpha,\beta)$ is the coordinate of the center, $\theta$ the angle between the radius at a point and the $X$ axis, measured counter-clockwise and $r$ the radius.

Now, in your new coordinate system, $(\alpha,\beta)\equiv (0,0)$ so you have the equations $$\frac{X_C}{\cos \theta} = \frac{Y_C}{\sin\theta} = r $$$$ \implies \cos \theta = {X_C\over r} \ \text{ and }\ \sin \theta = {Y_C\over r} \tag{1}$$ and $$\frac{X_B}{\cos (\theta+A)} = \frac{Y_B}{\sin(\theta+A)} = r\tag{2}$$

where you can find $\cos (\theta +A)$ and $\sin (\theta +A)$ using the compound angle formulae and $\sin \theta$ and $\cos \theta$ from $(1)$. Hence, you can find $(X_B, Y_B)$


Once you've found out $(X_B, Y_B)$ using the method outlined above, transform them back into your old coordinate system using $$x_B=X_B+h\\ y_B = Y_B+k$$

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Let $\sigma $ be angle made by BC to x-axis.

$$BC= 2 r \sin( A/2)$$ $$ \sigma =\pi/2 +A/2-\theta_B $$

$$x_C=x_B+ BC \cos \sigma $$ $$y_C=y_B- BC \sin \sigma . $$

No need to know origin.

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