Let two Circles of Radii
and
and centered at
and
intersect
in a Lens-shaped region. The equations of the two circles are
Combining (1) and (2) gives
![\begin{displaymath}
(x-d)^2+(R^2-x^2)=r^2.
\end{displaymath}](c1_1801.gif) |
(3) |
Multiplying through and rearranging gives
![\begin{displaymath}
x^2-2dx+d^2-x^2=r^2-R^2.
\end{displaymath}](c1_1802.gif) |
(4) |
Solving for
results in
![\begin{displaymath}
x={d^2-r^2+R^2\over 2d}.
\end{displaymath}](c1_1803.gif) |
(5) |
The line connecting the cusps of the Lens therefore has half-length given by plugging
back in to obtain
giving a length of
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}](c1_1811.gif) |
(8) |
twice, one for each half of the ``Lens.'' Noting that the heights of the two segment triangles are
The result is
The limiting cases of this expression can be checked to give 0 when
and
when
, as expected. In order for half the area of two Unit Disks (
) to overlap, set
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}](c1_1824.gif) |
(14) |
and solve numerically, yielding
.
See also Lens, Segment, Sphere-Sphere Intersection
© 1996-9 Eric W. Weisstein
1999-05-26