How To Find The Intersection Of Two Circles

The intersection surface area of ii circles

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Posted by Diego Assencio on 2017.07.12 under Mathematics (Geometry)

Let $C_1$ and $C_2$ be two circles of radii $r_1$ and $r_2$ respectively whose centers are at a distance $d$ from each other. Assume, without loss of generality, that $r_1 \geq r_2$. What is the intersection area of these 2 circles?

If $d \geq r_1 + r_2$, the circles intersect at almost upward to a point (when $d = r_1 + r_2$) and therefore the intersection area is zero. On the other farthermost, if $d + r_2 \leq r_1$, circle $C_2$ is entirely contained within $C_1$ and the intersection expanse is the expanse of $C_2$ itself: $\pi r_2^2$. The challenging case happens when both $d \lt r_1 + r_2$ and $d + r_2 \gt r_1$ are satisfied, i.eastward., when the the circles intersect only partially but the intersection surface area is more than than simply a point. Rearranging the second inequality, nosotros obtain $r_1 - r_2 \lt d \lt r_1 + r_2$, then we will assume this to be the case from now on.

To solve this problem, nosotros volition make use of a Cartesian coordinate system with origin at the heart of circle $C_1$ such that the center of $C_2$ is at $(d,0)$ as shown on effigy 1.

Fig. ane:

Two intersecting circles $C_1$ (bluish) and $C_2$ (ruby) of radii $r_1$ and $r_2$ respectively. The distance between the centers of the circles is $d = d_1 + d_2$, where $d_1$ is the $10$ coordinate of the intersection points and $d_2 = d - d_1$. Notice that $d_1 \geq 0$ since these points are always located to the right of the center of $C_1$, but $d_2$ may be negative when $r_2 \lt r_1$ since, in this case, the intersection points will somewhen fall to the right of the centre of $C_2$ as we move $C_2$ to the left, making $d \lt d_1$ and therefore $d_2 \lt 0$.

The circles $C_1$ and $C_2$ are described by the following equations respectively: $$ \begin{eqnarray} ten^ii + y^2 &=& r_1^2 \label{post_8d6ca3d82151bad815f78addf9b5c1c6_c1}\\[5pt] (x - d)^ii + y^ii &=& r_2^2 \\[5pt] \end{eqnarray} $$ At the intersection points, we have $x = d_1$. To decide $d_1$, we can supercede $x$ with $d_1$ and isolate $y^2$ on both equations above to get: $$ r_1^2 - d_1^two = r_2^2 - (d_1 - d)^two $$ Solving for $d_1$ is a simple job: $$ r_1^ii - d_1^2 = r_2^ii - d_1^ii + 2d_1d - d^2 \Longrightarrow d_1 = \displaystyle\frac{r_1^two - r_2^2 + d^ii}{2d} \label{post_8d6ca3d82151bad815f78addf9b5c1c6_eq_d1} $$ From equation \eqref{post_8d6ca3d82151bad815f78addf9b5c1c6_eq_d1}, nosotros can come across that $d_1 \geq 0$ since $r_1 \geq r_2$. The intersection area is the sum of the blueish and red areas shown on figure 1, which we refer to as $A_1$ and $A_2$ respectively. Nosotros then accept that: $$ \begin{eqnarray} A_1 &=& ii\int_{d_1}^{r_1} \sqrt{r_1^ii - x^2}dx \label{%INDEX_eq_A1_def} \\[5pt] A_2 &=& 2\int_{d - r_2}^{d_1} \sqrt{r_2^2 - (ten - d)^two}dx \end{eqnarray} $$ where the factors of $2$ come from the fact that each integral in a higher place accounts for merely one-half of the surface area of the associated region (merely points on and in a higher place the $ten$ axis are taken into account); the results must so exist multiplied by two and so that the areas below the $10$ centrality are taken into account as well.

The computation of these integrals is straightforward. Before nosotros proceed, notice offset that: $$ \brainstorm{eqnarray} A_2 &=& two\int_{d - r_2}^{d_1} \sqrt{r_2^2 - (x - d)^ii}dx \nonumber \\[5pt] &=& two\int_{- r_2}^{d_1 - d} \sqrt{r_2^2 - x^2}dx \nonumber \\[5pt] &=& ii\int_{d - d_1}^{r_2} \sqrt{r_2^2 - x^2}dx \nonumber \\[5pt] &=& 2\int_{d_2}^{r_2} \sqrt{r_2^ii - x^2}dx \characterization{%INDEX_eq_A2} \end{eqnarray} $$ where above we used the fact that $d_2 = d - d_1$. This is the aforementioned as equation \eqref{%INDEX_eq_A1_def} if we apply the substitutions $d_1 \rightarrow d_2$ and $r_1 \rightarrow r_2$. Therefore, past computing $A_1$, we will immediately obtain $A_2$ equally well. Let'southward and so compute $A_1$ first: $$ \begin{eqnarray} A_1 &=& ii\int_{d_1}^{r_1} \sqrt{r_1^ii - ten^2}dx \nonumber\\[5pt] &=& 2r_1 \int_{d_1}^{r_1} \sqrt{ane - \left(\frac{x}{r_1}\right)^ii}dx \nonumber\\[5pt] &=& 2r_1^2 \int_{d_1/r_1}^{1} \sqrt{one - x^2}dx \characterization{%INDEX_eq_A1} \end{eqnarray} $$ All we need to do now is to integrate $\sqrt{one - 10^2}$. The process is straightforward if nosotros employ integration by parts: $$ \begin{eqnarray} \int \sqrt{1 - x^two}dx &=& x \sqrt{1 - ten^ii} - \int x \left(\frac{-ten}{\sqrt{1 - ten^2}}\right) dx \nonumber\\[5pt] &=& 10 \sqrt{1 - 10^2} + \int \frac{x^2 - 1}{\sqrt{1 - x^2}} dx + \int \frac{1}{\sqrt{1 - x^2}} dx \nonumber\\[5pt] &=& x \sqrt{one - x^ii} - \int \sqrt{1 - x^2} dx + \sin^{-ane}(ten) \end{eqnarray} $$ Therefore: $$ \int \sqrt{1 - x^ii}dx = \frac{ane}{two}\left( ten \sqrt{1 - x^2} + \sin^{-1}(x) \right) \characterization{post_8d6ca3d82151bad815f78addf9b5c1c6_int_for_A1_A2} $$ Using equation \eqref{post_8d6ca3d82151bad815f78addf9b5c1c6_int_for_A1_A2} on equation \eqref{%INDEX_eq_A1} yields: $$ \brainstorm{eqnarray} A_1 &=& r_1^two \left( \frac{\pi}{2} - \frac{d_1}{r_1}\sqrt{1 - \left(\frac{d_1}{r_1}\correct)^2} - \sin^{-i}\left(\frac{d_1}{r_1}\right) \right) \nonumber\\[5pt] &=& r_1^2 \left( \cos^{-ane}\left(\frac{d_1}{r_1}\right) - \frac{d_1}{r_1}\sqrt{1 - \left(\frac{d_1}{r_1}\correct)^2} \right) \nonumber\\[5pt] &=& r_1^2 \cos^{-one}\left(\frac{d_1}{r_1}\right) - d_1 \sqrt{r_1^2 - d_1^2} \characterization{post_8d6ca3d82151bad815f78addf9b5c1c6_eq_A1_final} \end{eqnarray} $$ where in a higher place we used the fact that $\pi/two - \sin^{-1}(\blastoff) = \cos^{-1}(\alpha)$ for whatever $\alpha$ in $[-1,1]$. This fact is easy to prove: $$ \cos\left(\frac{\pi}{2} - \sin^{-1}(\alpha)\right) = \cos\left(\frac{\pi}{2}\correct)\cos(\sin^{-1}(\alpha)) + \sin\left(\frac{\pi}{2}\right)\sin(\sin^{-1}(\alpha)) = \alpha $$ and therefore $\pi/2 - \sin^{-1}(\alpha) = \cos^{-one}(\alpha)$. As discussed above, nosotros tin can now obtain $A_2$ directly by doing the substitutions $d_1 \rightarrow d_2$ and $r_1 \rightarrow r_2$ on the expression for $A_1$ on equation \eqref{post_8d6ca3d82151bad815f78addf9b5c1c6_eq_A1_final}: $$ A_2 = r_2^2 \cos^{-1}\left(\frac{d_2}{r_2}\right) - d_2 \sqrt{r_2^two - d_2^2} $$ The sum of $A_1$ and $A_2$ is the intersection area of the circles: $$ \boxed{ \brainstorm{eqnarray} A_{\textrm{intersection}} &=& r_1^ii \cos^{-1}\left(\frac{d_1}{r_1}\right) - d_1\sqrt{r_1^2 - d_1^ii} \nonumber \\[5pt] &+& r_2^two\cos^{-1}\left(\frac{d_2}{r_2}\right) - d_2\sqrt{r_2^2 - d_2^two} \nonumber \finish{eqnarray} } \label{post_8d6ca3d82151bad815f78addf9b5c1c6_A_intersection} $$ where: $$ \boxed{ d_1 = \displaystyle\frac{r_1^2 - r_2^two + d^ii}{2d} } \quad \textrm{ and } \quad \boxed{ d_2 = d - d_1 = \displaystyle\frac{r_2^2 - r_1^ii + d^2}{2d} } \label{post_8d6ca3d82151bad815f78addf9b5c1c6_eq_d1_final} $$

Summary

Given two circles $C_1$ and $C_2$ of radii $r_1$ and $r_2$ respectively (with $r_1 \geq r_2$) whose center points are at a altitude $d$ from each other, the intersection area of the circles is:

1.	zero, if $d \geq r_1 + r_2$, since in this case the circles intersect at most up to a point.
two.	$\pi r_2^ii$, if $d \leq r_1 - r_2$, since in this case $C_2$ is entirely contained inside $C_1$.
3.	given by equation \eqref{post_8d6ca3d82151bad815f78addf9b5c1c6_A_intersection} in all other cases.

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