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Circle-Circle Intersection


Let two Circles of Radii $R$ and $r$ and centered at $(0,0)$ and $(d,0)$ 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
\end{displaymath} (3)

Multiplying through and rearranging gives
\end{displaymath} (4)

Solving for $x$ results in
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 $x$ 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 $R'$and triangular height $d'$

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 $d=R+r$ 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 $r=R$, as expected. In order for half the area of two Unit Disks ($R=1$) to overlap, set $A=\pi R^2/2=\pi/2$ in the above equation
{\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

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© 1996-9 Eric W. Weisstein