Circle-Circle Intersection

$\begin{figure}\begin{center}\BoxedEPSF{CircleCircleIntersection.epsf}\end{center}\end{figure}$

Let two Circles of Radii and and centered at and intersect in a Lens-shaped region. The equations of the two circles are

$\displaystyle x^2+y^2$	$\textstyle =$	$\displaystyle R^2$	(1)
$\displaystyle (x-d)^2+y^2$	$\textstyle =$	$\displaystyle r^2.$	(2)

Combining (1) and (2) gives

$\begin{displaymath} (x-d)^2+(R^2-x^2)=r^2. \end{displaymath}$

(3)

Multiplying through and rearranging gives

$\begin{displaymath} x^2-2dx+d^2-x^2=r^2-R^2. \end{displaymath}$

(4)

Solving for

results in

$\begin{displaymath} x={d^2-r^2+R^2\over 2d}. \end{displaymath}$

(5)

The line connecting the cusps of the Lens therefore has half-length given by plugging

back in to obtain

$\displaystyle y^2$	$\textstyle =$	$\displaystyle R^2-x^2=R^2-\left({d^2-r^2+R^2\over 2d}\right)^2$
	$\textstyle =$	$\displaystyle {4d^2R^2-(d^2-r^2+R^2)^2\over 4d^2},$	(6)

giving a length of

$\displaystyle a$	$\textstyle =$	$\displaystyle {1\over d}\sqrt{4d^2R^2-(d^2-r^2+R^2)^2}$
	$\textstyle =$	$\displaystyle {1\over d}[(-d + r - R) (-d - r + R)[(-d + r + R) (d + r + R)]^{1/2}.$	(7)

This same formulation applies directly to the Sphere-Sphere Intersection problem.

To find the Area of the asymmetric ``Lens'' in which the Circles intersect, simply use the formula for the circular Segment of radius and triangular height

$\begin{displaymath} A(R',d')=R'^2\cos^{-1}\left({d'\over R'}\right)-d'\sqrt{R'^2-d'^2} \end{displaymath}$

(8)

twice, one for each half of the ``Lens.'' Noting that the heights of the two segment triangles are

$\displaystyle d_1$	$\textstyle =$	$\displaystyle x={d^2-r^2+R^2\over 2d}$	(9)
$\displaystyle d_2$	$\textstyle =$	$\displaystyle d-x={d^2+r^2-R^2\over 2d}.$	(10)

The result is

$\displaystyle A$	$\textstyle =$	$\displaystyle A(R_1, d_1)+A(R_2,d_2)$
	$\textstyle =$	$\displaystyle r^2\cos^{-1}\left({d^2+r^2-R^2\over 2dr}\right)+R^2\cos^{-1}\left({d^2+R^2-r^2\over 2dR}\right)$
	$\textstyle \phantom{=}$	$\displaystyle -{\textstyle{1\over 2}}\sqrt{(d-r-R)(d+r-R)(d-r+R)(d+r+R)}.$	(11)

The limiting cases of this expression can be checked to give 0 when

and

$\displaystyle A$	$\textstyle =$	$\displaystyle 2R^2\cos^{-1}\left({d\over 2R}\right)-{\textstyle{1\over 2}}d\sqrt{4R^2-d^2}$	(12)
	$\textstyle =$	$\displaystyle 2A({\textstyle{1\over 2}}d, R)$	(13)

when

, as expected. In order for half the area of two Unit Disks (

) to overlap, set $A=\pi R^2/2=\pi/2$ in the above equation

$\begin{displaymath} {\textstyle{1\over 2}}\pi=2\cos^{-1}({\textstyle{1\over 2}}d)-{\textstyle{1\over 2}}d\sqrt{4-d^2} \end{displaymath}$

(14)

and solve numerically, yielding $d\approx 0.807946$ .

See also Lens, Segment, Sphere-Sphere Intersection