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So I have a circle (I know its center's coordinates and radius) and a point on the circle (I know its coordinates) and I have to determine the coordinates of another point on the circle which is exactly at the distance L from the first point.

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What did you try? – Harald Hanche-Olsen Feb 24 '13 at 8:45

If I have undresttod corectly, assuming the equation of the circle as $$(x-x_0)^2+(y-y_0)^2=r^2$$ the fixed point $(x_1,y_1)$ and the desired point $(X,Y)$ then you need to solve the following equations simultaneously: $$(X-x_0)^2+(Y-y_0)^2=r^2$$ and $$(X-x_1)^2+(Y-y_1)^2=L^2$$

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Further suggestion: Expand the squares in both equations and subtract them from another to get a linear equation in $X$ and $Y$. Use it to eliminate $Y$ from one of the quadratic equations. – Harald Hanche-Olsen Feb 24 '13 at 8:50
+1 I think you "nailed it", and with the suggestion from @Harald - the OP is on his/her way! – amWhy Feb 24 '13 at 13:36

If the equation of the circle be $(x-h)^2+(y-k)^2=r^2$ any point on the circle can be parametrized as $(r\cos\theta+h,r\sin\theta+k)$ and if the known point on the circle be $S(p,q)$

So, we need to know the two values of $\theta$ from $(r\cos\theta+h-p)^2+(r\sin\theta+k-q)=L^2$

$\implies 2r(h-p)\cos\theta+2r(k-q)\sin\theta=L^2-r^2-(h-p)^2-(k-q)^2$

Now, we know $-\sqrt{c^2+d^2}\le c\cos A+d\sin A\le \sqrt{c^2+d^2}$(Proof below)

There will be two or zero solutions according as $c\cos A+d\sin A$ lies within or outside the range.


Proof: Let $c=R\cos B,d=R\sin B$ where real $R\ge0$

So, $c^2+d^2=R^2\implies R=\sqrt{c^2+d^2}$ and $\tan B=\frac dc$

and $c\cos A+d\sin A=\sqrt{c^2+d^2}\cos(A-B)$

As $-1\le\cos(A-B)\le1, -\sqrt{c^2+d^2}\le\sqrt{c^2+d^2}\cos(A-B)\le \sqrt{c^2+d^2}$

$\implies -\sqrt{c^2+d^2}\le \cos A+d\sin A\le \sqrt{c^2+d^2}$


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