Distance between points on two circles

[LucKy]

Hello all

I summarised my question on the attached picture

I will wery appreciate if somebody can confirm:
Does my attempt to find distance a2_b2 correct? Is there simplier method?

It is not a homework, It is for my hobby project.
(I want to simulate my robot's sensor readings in some situations)

My attempt:

Step 1. Calculate ${\angle \beta}$:
Assume ${\angle \beta}$ is directly proportional to ${\angle \alpha}$ (NOT SURE!)

$$ \frac {\angle \beta} {\angle \alpha} = \frac {\angle b1\_O2\_b3 } {\angle a1\_O1\_a3} $$
$$ {\angle b1\_O2\_b3 }= \frac \pi 2 \text{ , } {\angle a1\_O1\_a3} = \tan^{-1}(\frac R D) $$
$$ {\angle \beta} = \frac {\angle \alpha * \frac \pi 2} {\tan^{-1}(\frac R D)} $$

Step2. Calculate distance O2_p:
$$ {O2\_p} = {R*\cos(\beta)}$$

Step3. Calculate distance O1_p
$$ {O1\_p} = {D + O2\_p} = {D + R*\cos(\beta)}$$

Step4. Calculate distance O1_b2
$$ {O1\_b2} = \frac {O1\_p} {\cos(\alpha)} = \frac {D + R*\cos(\beta)} {\cos(\alpha)}$$

Step5. (Final) Calculate distance a2_b2
$$ {a2\_b2} = {O1\_b2 - R} = \frac {D + R*\cos(\beta)} {\cos(\alpha)} - R \text{, where } {\angle \beta} = \frac {\angle \alpha * \frac \pi 2} {\tan^{-1}(\frac R D)} $$

 

[kuruman]

Is there simplier method?

Note that
$y=R\sin\beta=x\sin\alpha.$ Angles $\alpha$ and $\beta$ are not proportional.

Then
$\cos\beta=\sqrt{1-\sin^2\beta}=\dfrac{1}{R}\sqrt{R^2-x^2\sin^2\alpha}.$

From the law of cosines for triangle ABC
$x^2=D^2+R^2+2DR\cos\beta \implies x^2=D^2+R^2+2DR\left(\dfrac{1}{R}\sqrt{R^2-x^2\sin^2\alpha}\right).$

Solve the quadratic for $u=x^2$. Make sure you pick the correct solution then subtract $R$ from $\sqrt{u}$.

 

[bob012345]

I don’t know if this is simpler but it can be done as the intersection of a line and a circle. Note, I’m using $x$ as the standard coordinate. Referring to @kuruman ’s excellent graphic, If the origin is at A and letting D be along the x axis, the point C is the intersection of the line $y=xtan(\alpha)$ and the circle centered at $(D,0)$ which is $(x-D)^2+y^2=R^2$. Solve for $x$ then $AC=\large\frac{x}{cos(\alpha)}$ then just subtract $R$.

 

[kuruman]

I don’t know if this is simpler . . .

If it isn't simpler, it certainly is more elegant.