1
$\begingroup$

graph

As in the image, I have a circle that intersect the $y$ axis (or could be a vertical line). I know center and radius of that circle. There is a second smaller circle, inside the bigger one, tangent to the $y$ axis (or vertical line) and to the bigger circle. I know the coordinate $y_A$ where the smaller circle intersect the vertical line. I have to calculate the radius of the smaller circle $r$. I try to do it using trigonometry, but I found a solution to be iterate. I'm looking for a "direct" solution.

$\endgroup$
2
  • 1
    $\begingroup$ You can draw a triangle with the three corners being $(0,y_a), (x_a, y_a), (x_C,y_c)$. The edge lengths of the triangle are $r, R-r$ and $d=\sqrt{x_C^2 + (y_C-y_a)^2}$. Now you can perhaps solve something from that ... $\endgroup$ – Matti P. Oct 22 '20 at 9:49
  • $\begingroup$ Let $\ell$ be the vertical line $x=R$. Then the distance from $A$ to $\ell$ equals the distance from $A$ to $C$, that is, the locus of valid $A$ is on a parabola. $\endgroup$ – Hagen von Eitzen Oct 22 '20 at 9:53
2
$\begingroup$

Obviously the circle's centre will be at $(r,y_A)$. Then we have the following relation from the tangency with the larger circle: $$\sqrt{(x_C-r)^2+(y_A-y_C)^2}=R-r$$ Let $(y_A-y_C)^2=K$, then squaring both sides: $$(x_C-r)^2+K=R^2-2Rr+r^2$$ $$x_C^2-2x_Cr+K=R^2-2Rr$$ $$r(2R-2x_C)=R^2-x_C^2-K$$ $$r=\frac{R^2-x_C^2-K}{2R-2x_C}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.