# Finding the intersection points of straight lines (from the intersection of outer circle radius and inner circle) and the outer circle

Given two circles with the same origin, and:

• their radii are $$r_1$$ and $$r_2$$
• radii difference is $$s=|r_2-r_1|$$
• $$(x_0,y_0)$$ is an arbitrary point on the inner circle (that is, on the one with smaller radius, though it's also possible that $$r_1=r_2$$)

Extending two straight lines (i.e. one with no slope and one with slope of $$0$$) from $$(x_0,y_0)$$ find where these intersect the outer circle. Let's call the closest intersection point of the vertical line $$(x_1,y_1)$$ (note that $$x_1=x_0$$) and of the horizontal line $$(x_2,y_2)$$ (note that $$y_2=y_0$$). Other intersection points can be gotten by simply rotating these around the axes, so you can just ignore them in this problem.

This is the best visual representation I could draw: here.

This is based on a practical problem I have but I think I've included all the relevant information.

Take a look at the following figure: (where I have assumed that $$r_1 < r_2$$, that the origin is in the common center, and $$a$$ the "main angle").

As point $$A$$ has coordinates $$\binom{r_1 \cos a}{r_1 \sin a}$$ and $$B$$ has coordinates $$\binom{r_2 \cos b}{r_2 \sin b}$$ and they have the same ordinates, we can deduce that :

$$r_1 \sin a=r_2 \sin b \ \iff \ b=\sin^{-1}\left(\frac{r_1}{r_2} \sin a\right)$$

Knowing $$b$$, you have the coordinates of $$B$$, and length

$$AB=r_2 \cos b-r_1 \cos a=r_2 \sqrt{1-\left(\frac{r_1}{r_2} \sin a\right)^2} -r_1 \cos a$$

The same kind of reasoning gives the coordinates of point $$C$$ belonging to the biggest circle and having the same abscissa (this time) as point $$A$$.

• Any comment ?... Jan 11 at 8:09
• I just added a formula for the computation of length $AB$. Jan 11 at 8:19
• That's it, thanks. 👍 Jan 12 at 13:26