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I am having trouble getting the $(x,y)$ of a certain point on the circle. Please look at the image:

The circles are the identical, the radius is $1000 \text{ units}$, $S$ is the center with coordinates on the top right. This is not homework just to be clear.

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One messy but guaranteed method will be finding the slope of line connecting $S_1$ and your point (which is perpendicular to $S_1 S_2$ then solving the simultaneous equation $(x-765)^2+(y-211)^2=1000^2$ and your equation of line. –  user160738 Jul 3 at 21:44
    
Hint: find the equation of the line $d(S_1,S_2)$, and than find the line normal to that line that passes through $S_1$. In the end find the point that is 1000 units far from $S_1$ (there are two such points, but the right one is that with the bigger y-coordinate). –  Emin Jul 3 at 21:45
    
I forgot to mention a catch: I have to do this on a computer, so writing equations would be complex. –  Uros Hercog Jul 3 at 21:48
    
The fact that $S_2$ is the centre of a circle is completely irrelevant here! It's just a point. –  TonyK Jul 3 at 22:27

2 Answers 2

Method using only equations: refer to my comment above.

One less messy method would be using vectors. The vector from $S_1$ to $S_2$ is $(767,212)-(765,211)=(2,1)$. Rotate this vector 90 degrees anticlockwise, to get one from $S_1$ to direction of your required point: $(-1,2)$. The unit vector in this direction is $(-1,2)\frac{1}{\sqrt{1+2^2}}$

Now your point is: $(765,211)+r(-1,2)\frac{1}{\sqrt{1+2^2}}$ where $r$ is the radius of circle

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You need a vector perdendicular to that line, i.e. dot product must be zero: $$\begin{pmatrix}x \\ y\end{pmatrix}\cdot \begin{pmatrix}x_2-x_1 \\ y_2-y_1\end{pmatrix} = 0$$ The other equation is that the length of your vector must be the radius $r$: $$x^2+y^2=r^2$$ Now you can put one equation in the other and solve it, yielding something like: $$x^2 = \frac{r^2(y_2-y_1)^2}{(x_2-x_1)^2+(y_2-y_1)^2}$$ $$y=\sqrt{r^2-x^2}$$ Denominator of the first equation should be positive as long the points $S_1,S_2$ are not the same point. Also, the first equation has two solutions, of which you must pick the one with the correct orientation. Anyway, you could enter these two equations in your computer to get the result (I hope they're correct, but better check it yourself one more time).

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