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November 23

radial distance between a circle and another enclosing circle

On an x-y plane, draw a circle, radius r1 centered on the origin, 0,0. Draw a second circle centered on some offset value -x, y = 0, radius r2 which greater than r1+x so that the second circle completely encloses the first and does not touch it. Draw a line at angle a beginning at the origin and crossing both circles. How do I calculate the distance along this line between the two circles? ```` Dionne Court (talk) 06:07, 23 November 2024 (UTC)[reply]

Given:

inner circle: centre at $(0,0),$ radius $r_{1},$ equation $x^{2}+y^{2}=r_{1}^{2};$
outer circle: centre at $(u,0),$ radius $r_{2}>r_{1}+u,$ equation $(x-u)^{2}+y^{2}=r_{2}^{2}.$
line through origin at angle $\alpha ,$ parametric equation $(x,y)=(\lambda \cos \alpha ,\lambda \sin \alpha ).$

The line crosses the inner circle at

(x,y)=(\pm r_{1}\cos \alpha ,\pm r_{1}\sin \alpha ),

both obviously at distance

r_{1}

from the origin.

To find its crossings with the outer circle, we substitute the rhs of the line's equation for

(x,y)

into the equation of the outer circle, giving

(\lambda \cos \alpha -u)^{2}+(\lambda \sin \alpha )^{2}=r_{2}^{2}.

We need to solve this for the unknown

\lambda

. This is a quadratic equation; call its roots

\lambda _{1}

and

\lambda _{2}.

The corresponding points are at distances

|\lambda _{1}|

and

|\lambda _{2}|

from the origin.

The crossing distances are then

|\lambda _{1}|-r_{1}

and

|\lambda _{2}|-r_{1}.

If you use

\left|(|\lambda _{1}|-r_{1})\right|

and

\left|(|\lambda _{1}|-r_{1})\right|,

this will work for any second circle, also of it intersects the origin-centred circle or is wholly inside, provided the quadratic equation has real-valued roots. --Lambiam 08:46, 23 November 2024 (UTC)[reply]