trigonometry
I am having trouble trying to work out how to calculate the $(x,y)$ point around a circle for a given distance from the circles center.
Variables I do have are:
constant distance/radius from center ($r$)
the angle from $y$ origin
I basically need a point ($x$ and $y$) around a circle every $18$ degrees.
Excuse me if this is a very basic question but math is not my strong point :/
Ta
John
Update 2: Here is the graph I got for $(x_{1},y_{1})=( 78. 965,12. 354)$, for the parametric circle $(x(t),y(t))$ centered at $(100,100)$
$$x=100+90.135\cos \left( 1.3527+\pi -t\frac{\pi }{180}\right) ,$$
$$y=100+90.135\sin \left( 1.3527+\pi -t\frac{\pi }{180}\right) .$$
together with the 4 points $(x(t),y(t))$ for $t=0,90,180,270$
$$(x_{1},y_{1})=(x(0),y(0)),(x(90),y(90)),(x(180),y(180)),(x(270),y(270)).$$
You might use the following equations in a for loop with $k=0$ to $k=359$, step $1$:
$$x=100+90.135\cos \left( 1.3527+\pi -k\frac{\pi }{180}\right) ,$$
$$y=100+90.135\sin \left( 1.3527+\pi -k\frac{\pi }{180}\right) .$$
to draw the "orbit" with a 1 degree interval.
Update: corrected coordinates of $(x_{1},y_{1})=(140.5,152)$.
You need to consider the new angle and not only the $1{{}^\circ}$ change. The argument of $\cos$ and $\sin$ is this new angle and not $1{{}^\circ}$.
Let $(x_{c},y_{c})=(160,240)$ be the center of the set of circles and $(x_{1},y_{1})=(140.5,152)$. The radius $r$ is
$$\begin{eqnarray*} r &=&\sqrt{\left( x_{c}-x_{1}\right) ^{2}+\left( y_{c}-y_{1}\right) ^{2}} \\ &=&\sqrt{\left( 160-140.5\right) ^{2}+\left( 240-152\right) ^{2}} \\ &=&90.135 \end{eqnarray*}$$
Call $(x,y)$ the new coordinates of $(x_{1},y_{1})$ rotated by an angle of $-1{{}^\circ}=-\dfrac{\pi }{180}$ around $(x_{c},y_{c})$ with a radius $r$. The new angle is $\theta'=\theta -\frac{\pi }{180}$, $\theta $ being the initial angle. Then
$$\begin{eqnarray*} x &=&x_{c}+r\cos \left( \theta -\frac{\pi }{180}\right), \\ y &=&y_{c}+r\sin \left( \theta -\frac{\pi }{180}\right), \end{eqnarray*}$$ where $\theta $ is the angle $\theta =\arctan \dfrac{y_{1}-y_{c}}{x_{1}-x_{c}}:$
$$\begin{eqnarray*} \theta &=&\arctan \frac{152-240}{140.5-160}=1.3527+\pi \text{ rad.}\\ &=&\frac{1.3527\times 180{{}^\circ}}{\pi }+180{{}^\circ}=257. 5{{}^\circ}\end{eqnarray*}$$ Thus $$\begin{eqnarray*} x &=&160+90.135\cos \left( 1.3527+\pi -\frac{\pi }{180}\right)= 138. 96 \\y &=&240+90.135\sin \left( 1.3527+\pi -\frac{\pi }{180}\right) = 152. 35 \end{eqnarray*}$$
Let $X$ be the external point. Let $P$ be the center of the circle. Let $Q$ be a point of tangency. Then triangle $\Delta PQX$ is a right triangle. If we let $\theta$ be the angle that $PX$ makes with the horizontal, then the angle of the tangent point with the horizontal is $\theta + \angle XPQ$ and $\theta - \angle XPQ$ respectively. Consequently, the angle that the tangents make with the vertical will also be $\theta \pm \angle XPQ$.
In the figure above, the green angle would be the angle give by $\theta + \angle XPQ$.
If $X = (x_2, y_2)$ and $P = (x_1, y_1)$ then $\theta = \arctan\left(\frac{y_2 - y_1}{x_2 - x_1}\right)$.
If you want the heading to the tangent point, then you will want the red angle (there is another angle for the lower tangent which I have not labelled, it is given analogously) which will be $90^{\circ} - (\theta - \angle XPQ)$.
If you want the angle the tangent makes with the vertical, you will want the blue angle (or the supplement of the blue angle) which is equal to the green angle, $\theta + \angle XPQ$.
You may have to tweak a bit to make sure that everything works out in all quadrants, but this is the basic idea.
Edit If you want $\angle FXQ$ as defined in your comment, then it is simple. Draw a vertical line intersecting $X$. Then the tangent cuts through the vertical line and the $y$ axis as parallel lines. $\angle FXQ$ is then given by the blue angle.
Best Answer
Inverse Y-axis:
If radians is used then
and
Radian is the standard unit of angular measure, any time you see angles, always assume they are using radians unless told otherwise.