Testwiki:Reference desk/Archives/Mathematics/2024 November 23

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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)

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 $|(|{\lambda }_1|-r_1)|$ and $|(|{\lambda }_1|-r_1)|,$ 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)