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I want to find the radius $R$ knowing

  • the coordinates of two points $(x_1,y_1)$ and $(x_2,y_2)$
  • the angle $\alpha$ between a tangent to the circle passing by $(x_1,y_1)$ and the horizontal

find R

Note that $x_1$<$x_2$ and $0\le\alpha<\pi$


There is a similar question here but I think it's possible to find a geometric answer that does not involve solving a system of equation

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The centre of the circle has coordinates $$(x_1+R\sin\alpha, y_1-R\cos\alpha)$$

If $P=(x_2, y_2)$, then considering the distance $CP^2$, $$(x_1-x_2)^2+2R(x_1-x_2)\sin\alpha+R^2\sin^2\alpha+(y_1-y_2)^2-2R(y_1-y_2)\cos\alpha+R^2\cos^2\alpha=R^2$$

Now you can cancel terms and rearrange to get $R$

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  • $\begingroup$ Just want to add an extra info. The equation thus set up will NOT be a quadratic. $\endgroup$ – Mick Jan 1 '16 at 16:28
  • $\begingroup$ @Mick| yes, the $R^2$ terms cancel $\endgroup$ – David Quinn Jan 1 '16 at 17:44

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