I am facing a problem that I have difficulty to solve.

I have a line that originates from the origin (point $(0,0)$) and has a known angle to the $x$ axis (angle $\theta$)

Somewhere between the $x$ axis and this line is a circle. This circle touches both lines, but the dimension and center of this circle are not known.

On the right side of the circle is a point $P$ that lies on the circumference of the circle and of which the coordinates are known.

How can I find the coordinates of the center of the circle?

I can't seem to add an image so I have used an external website to make the drawing.


I now realize that I have over-simplified the problem because it is hard to explain in words. However, now I know that I can send a drawing in another way I would like to post the original problem:

You can see the drawing over here

There are two lines originating from the origin. The angle of each line in relation to the diagonal is the same and is known. A circle is centered on the diagonal and only a 90 degrees arc of the circle is drawn between the lines. On the arc of the circle is a point P of which the coordinates are known. The dimension and location of the circle is not known, the only thing that is known is that point P lies on the arc and the arc is a 90 degrees segment of a circle with its center on the diagonal.

How can I find the coordinates of the circle?

  • 1
    $\begingroup$ A drawing would help a lot. There seems to be enough information in the text, but it is hard to be sure. $\endgroup$ – Ross Millikan Oct 31 '20 at 2:13
  • $\begingroup$ With the new changes, you should consider posting a new problem instead $\endgroup$ – Andrei Oct 31 '20 at 5:27

A drawing of my understanding is below. Please make sure it matches your understanding. We know $\theta$ and the coordinates of $P$. Because the circle is tangent to the two lines, $D$ is on the bisector of the angle. That gives you the slope of the line from the origin through $D$, which gives a relationship between the $x$ and $y$ coordinates of $D$. You can now write the law of cosines for the triangle $ADP$ in terms of the $x$ coordinate of $D$ and solve the equation.

enter image description here

  • $\begingroup$ Thank you for the quick repley. However I can't visit i.stack.imgur.com - my browser says there is a problem with the site - so I can't see the image you have attached. $\endgroup$ – A. vanleyen Oct 31 '20 at 4:15
  • $\begingroup$ I have no problem. Perhaps a cache clear is needed $\endgroup$ – Ross Millikan Oct 31 '20 at 4:19
  • $\begingroup$ I think it is because I am living in Shenzhen which is in China.. I have however used an external website (desmos) to make a drawing that you can view: $\endgroup$ – A. vanleyen Oct 31 '20 at 4:21
  • $\begingroup$ I have however used an external website (desmos) to make a drawing that you can view :desmos.com/geometry/gq8xv1yumt $\endgroup$ – A. vanleyen Oct 31 '20 at 4:24
  • $\begingroup$ Yes, that matches mine nicely. My $D$ is the center of the circle. $\endgroup$ – Ross Millikan Oct 31 '20 at 4:34

The equation of the line at angle $\theta$ is $$y=x\tan\theta$$ Let's call the coordinates of the center of the circle $(x_c,y_c)$. Then you know that the distance from the center to the line is equal to the distance from the center to point $P$ (note that this is on a parabola with focus $P$, and the directrix is the tilted line). We write this as $$\frac{(x_c\tan\theta-y_c)^2}{\tan^2\theta+1}=(x_c-x_P)^2+(y_c-y_P)^2$$ I've used the formula on this link, but you can just search for it. Since the circle touches the line and the $x$ axis, it means that the center is on the bisector of the angle. Therefore $$\tan\frac\theta 2=\frac{y_c}{x_c}$$ Take $y_c$ from the second equation, plug it into the first one, and you just get a quadratic. One of the solution corresponds to $P$ on the right side of the circle, one on the left. In your case, you want the smaller number.

  • $\begingroup$ Ross and Andrei, thank you very much. $\endgroup$ – A. vanleyen Oct 31 '20 at 4:58
  • $\begingroup$ However, the problem is more complicated as I originally said. Pleas see my modified original post $\endgroup$ – A. vanleyen Oct 31 '20 at 4:59

For the modified problem, draw a vertical line through $C$. Since $C$ is on the diagonal, $x_C=y_C$. That's one of the unknowns. The second is the radius $r$. So you need two equations. One of them is the distance to point $P$: $$(x_P-x_C)^2+(y_P-x_C)^2=r^2$$ For the other equation, look at the triangle formed by the $x$ axis, the upper line, and the vertical through $C$. This is a right angle triangle, with the angle at origin $\pi/4+\theta$, the opposite side is $x_C+r$, and the adjacent side $x_C$. Then the other equation is: $$\tan\left(\frac\pi4+\theta\right)=\frac{x_C+r}{x_C}$$ Eliminating $r$ from these two equations you get a quadratic in $x_C$. You need to choose the solution where $x_C<x_P$.


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