Ingeometry, anintersection is a point, line, or curve common to two or more objects (such as lines, curves, planes, and surfaces). The simplest case inEuclidean geometry is theline–line intersection between two distinctlines, which either is onepoint (sometimes called avertex) or does not exist (if the lines areparallel). Other types of geometric intersection include:

• Line–plane intersection
• Line–sphere intersection
• Intersection of a polyhedron with a line
• Line segment intersection
• Intersection curve

Determination of the intersection offlats – linear geometric objects embedded in a higher-dimensional space – is a simple task oflinear algebra, namely the solution of asystem of linear equations. In general the determination of an intersection leads tonon-linear equations, which can besolved numerically, for example usingNewton iteration. Intersection problems between a line and aconic section (circle, ellipse, parabola, etc.) or aquadric (sphere, cylinder, hyperboloid, etc.) lead toquadratic equations that can be easily solved. Intersections between quadrics lead toquartic equations that can be solvedalgebraically.

For the determination of the intersection point of two non-parallel lines

${\displaystyle a_{1}x+b_{1}y=c_1,a_{2}x+b_{2}y=c_2}$

one gets, fromCramer's rule or by substituting out a variable, the coordinates of the intersection point${\displaystyle (x_s,y_s)}$ :

${\displaystyle x_s=\frac{c_1{b}_2-c_2{b}_1}{a_1{b}_2-a_2{b}_1},y_s=\frac{a_1{c}_2-a_2{c}_1}{a_1{b}_2-a_2{b}_1}.}$

(If${\displaystyle a_1{b}_2-a_2{b}_1=0}$ the lines are parallel and these formulas cannot be used because they involve dividing by 0.)

For two non-parallelline segments${\displaystyle (x_1,y_1),(x_2,y_2)}$ and${\displaystyle (x_3,y_3),(x_4,y_4)}$ there is not necessarily an intersection point (see diagram), because the intersection point${\displaystyle (x_0,y_0)}$ of the corresponding lines need not to be contained in the line segments. In order to check the situation one uses parametric representations of the lines:

${\displaystyle (x(s),y(s))=(x_1+s(x_2-x_1),y_1+s(y_2-y_1)),}$

${\displaystyle (x(t),y(t))=(x_3+t(x_4-x_3),y_3+t(y_4-y_3)).}$

The line segments intersect only in a common point${\displaystyle (x_0,y_0)}$ of the corresponding lines if the corresponding parameters${\displaystyle s_0,t_0}$ fulfill the condition${\displaystyle 0\le s_0,t_0\le 1}$. The parameters${\displaystyle s_0,t_0}$ are the solution of the linear system

${\displaystyle s(x_2-x_1)-t(x_4-x_3)=x_3-x_1,}$

${\displaystyle s(y_2-y_1)-t(y_4-y_3)=y_3-y_1.}$

It can be solved fors andt using Cramer's rule (seeabove). If the condition${\displaystyle 0\le s_0,t_0\le 1}$ is fulfilled one inserts${\displaystyle s_0}$ or${\displaystyle t_0}$ into the corresponding parametric representation and gets the intersection point${\displaystyle (x_0,y_0)}$.

Example: For the line segments${\displaystyle (1,1),(3,2)}$ and${\displaystyle (1,4),(2,-1)}$ one gets the linear system

${\displaystyle 2s-t=0}$

${\displaystyle s+5t=3}$

and${\displaystyle s_0=\frac{3}{11},t_0=\frac{6}{11}}$. That means: the lines intersect at point${\displaystyle (\frac{17}{11},\frac{14}{11})}$.

Remark: Considering lines, instead of segments, determined by pairs of points, each condition${\displaystyle 0\le s_0,t_0\le 1}$ can be dropped and the method yields the intersection point of the lines (seeabove).

For the intersection of

• line${\displaystyle ax+by=c}$ andcircle${\displaystyle x^2+y^2=r^2}$

one solves the line equation forx ory andsubstitutes it into the equation of the circle and gets for the solution (using the formula of a quadratic equation)${\displaystyle (x_1,y_1),(x_2,y_2)}$ with

${\displaystyle x_{1/2}=\frac{ac\pm b\sqrt{r^2(a^2+b^2)-c^2}}{a^2+b^2},}$

${\displaystyle y_{1/2}=\frac{bc\mp a\sqrt{r^2(a^2+b^2)-c^2}}{a^2+b^2},}$

if${\displaystyle r^2(a^2+b^2)-c^2>0.}$ If this condition holds with strict inequality, there are two intersection points; in this case the line is called asecant line of the circle, and the line segment connecting the intersection points is called achord of the circle.

If${\displaystyle r^2(a^2+b^2)-c^2=0}$ holds, there exists only one intersection point and the line is tangent to the circle. If the weak inequality does not hold, the line does not intersect the circle.

If the circle's midpoint is not the origin, see.^[1] The intersection of a line and a parabola or hyperbola may be treated analogously.

The determination of the intersection points of two circles

• ${\displaystyle (x-x_1{)}^2+(y-y_1{)}^2=r_1^2,(x-x_2{)}^2+(y-y_2{)}^2=r_2^2}$

can be reduced to the previous case of intersecting a line and a circle. By subtraction of the two given equations one gets the line equation:

${\displaystyle 2(x_2-x_1)x+2(y_2-y_1)y=r_1^2-x_1^2-y_1^2-r_2^2+x_2^2+y_2^2.}$

This special line is theradical line of the two circles.

Special case${\displaystyle x_1=y_1=y_2=0}$ :
In this case the origin is the center of the first circle and the second center lies on the x-axis (s. diagram). The equation of the radical line simplifies to${\displaystyle 2x_{2}x=r_1^2-r_2^2+x_2^2}$ and the points of intersection can be written as${\displaystyle (x_0,\pm y_0)}$ with

${\displaystyle x_0=\frac{r_1^2-r_2^2+x_2^2}{2x_2},y_0=\sqrt{r_1^2-x_0^2}.}$

In case of the circles have no points in common.
In case of${\displaystyle r_1^2=x_0^2}$ the circles have one point in common and the radical line is a common tangent.

Any general case as written above can be transformed by a shift and a rotation into the special case.

The intersection of twodisks (the interiors of the two circles) forms a shape called alens.

The problem of intersection of an ellipse/hyperbola/parabola with anotherconic section leads to asystem of quadratic equations, which can be solved in special cases easily by elimination of one coordinate. Special properties of conic sections may be used to obtain asolution. In general the intersection points can be determined by solving the equation by a Newton iteration. If a) both conics are given implicitly (by an equation) a 2-dimensional Newton iteration b) one implicitly and the other parametrically given a 1-dimensional Newton iteration is necessary. See next section.

Two curves in${\displaystyle {\mathbb{R}}^2}$ (two-dimensional space), which are continuously differentiable (i.e. there is no sharp bend),have an intersection point, if they have a point of the plane in common and have at this point (see diagram):

a) different tangent lines (transversal intersection, aftertransversality), or

b) the tangent line in common and they are crossing each other (touching intersection, aftertangency).

If both the curves have a pointS and the tangent line there in common but do not cross each other, they are justtouching at pointS.

Because touching intersections appear rarely and are difficult to deal with, the following considerations omit this case. In any case below all necessary differential conditions are presupposed. The determination of intersection points always leads to one or two non-linear equations which can be solved by Newton iteration. A list of the appearing cases follows:

• Ifboth curves are explicitly given:${\displaystyle y=f_1(x),y=f_2(x)}$, equating them yields the equation

${\displaystyle f_1(x)=f_2(x).}$

• Ifboth curves are parametrically given:${\displaystyle C_1:(x_1(t),y_1(t)),C_2:(x_2(s),y_2(s)).}$

Equating them yields two equations in two variables:

${\displaystyle x_1(t)=x_2(s),y_1(t)=y_2(s).}$

• Ifone curve is parametrically and the other implicitly given:${\displaystyle C_1:(x_1(t),y_1(t)),C_2:f(x,y)=0.}$

This is the simplest case besides the explicit case. One has to insert the parametric representation of${\displaystyle C_1}$ into the equation${\displaystyle f(x,y)=0}$ of curve${\displaystyle C_2}$ and one gets the equation:

${\displaystyle f(x_1(t),y_2(t))=0.}$

• Ifboth curves are implicitly given:${\displaystyle C_1:f_1(x,y)=0,C_2:f_2(x,y)=0.}$

Here, an intersection point is a solution of the system

${\displaystyle f_1(x,y)=0,f_2(x,y)=0.}$

Any Newton iteration needs convenient starting values, which can be derived by a visualization of both the curves. A parametrically or explicitly given curve can easily be visualized, because to any parametert orx respectively it is easy to calculate the corresponding point. For implicitly given curves this task is not as easy. In this case one has to determine a curve point with help of starting values and an iteration. See.^[2]

Examples:

1:${\displaystyle C_1:(t,t^3)}$ and circle${\displaystyle C_2:(x-1{)}^2+(y-1{)}^2-10=0}$ (see diagram).

The Newton iteration${\displaystyle t_{n+1}:=t_n-\frac{f(t_n)}{f^{\prime }(t_n)}}$ for function

${\displaystyle f(t)=(t-1{)}^2+(t^3-1{)}^2-10}$ has to be done. As start values one can choose −1 and 1.5.

The intersection points are: (−1.1073, −1.3578), (1.6011, 4.1046)

2:${\displaystyle C_1:f_1(x,y)=x^4+y^4-1=0,}$

${\displaystyle C_2:f_2(x,y)=(x-0.5{)}^2+(y-0.5{)}^2-1=0}$ (see diagram).

The Newton iteration

${\displaystyle (\genfrac{}{}{0ex}{}{x_{n+1}}{y_{n+1}})=(\genfrac{}{}{0ex}{}{x_n+{\delta }_x}{y_n+{\delta }_y})}$ has to be performed, where${\displaystyle (\genfrac{}{}{0ex}{}{{\delta }_x}{{\delta }_y})}$ is the solution of the linear system

at point${\displaystyle (x_n,y_n)}$. As starting values one can choose(−0.5, 1) and (1, −0.5).

The linear system can be solved by Cramer's rule.

The intersection points are (−0.3686, 0.9953) and (0.9953, −0.3686).

If one wants to determine the intersection points of twopolygons, one can check the intersection of any pair of line segments of the polygons (seeabove). For polygons with many segments this method is rather time-consuming. In practice one accelerates the intersection algorithm by usingwindow tests. In this case one divides the polygons into small sub-polygons and determines the smallest window (rectangle with sides parallel to the coordinate axes) for any sub-polygon. Before starting the time-consuming determination of the intersection point of two line segments any pair of windows is tested for common points. See.^[3]

In 3-dimensional space there are intersection points (common points) between curves and surfaces. In the following sections we considertransversal intersection only.

The intersection of a line and a planeingeneral position in three dimensions is a point.

Commonly a line in space is represented parametrically${\displaystyle (x(t),y(t),z(t))}$ and a plane by an equation${\displaystyle ax+by+cz=d}$. Inserting the parameter representation into the equation yields the linear equation

${\displaystyle ax(t)+by(t)+cz(t)=d,}$

for parameter${\displaystyle t_0}$ of the intersection point${\displaystyle (x(t_0),y(t_0),z(t_0))}$.

If the linear equation has no solution, the line either lies on the plane or is parallel to it.

If a line is defined by two intersecting planes${\displaystyle {\epsilon }_i:{\overrightarrow{n}}_i\cdot \overrightarrow{x}=d_i,i=1,2}$ and should be intersected by a third plane${\displaystyle {\epsilon }_3:{\overrightarrow{n}}_3\cdot \overrightarrow{x}=d_3}$, the common intersection point of the three planes has to be evaluated.

Three planes${\displaystyle {\epsilon }_i:{\overrightarrow{n}}_i\cdot \overrightarrow{x}=d_i,i=1,2,3}$ with linear independent normal vectors${\displaystyle {\overrightarrow{n}}_1,{\overrightarrow{n}}_2,{\overrightarrow{n}}_3}$ have the intersection point

${\displaystyle {\overrightarrow{p}}_0=\frac{d_1({\overrightarrow{n}}_2\times {\overrightarrow{n}}_3)+d_2({\overrightarrow{n}}_3\times {\overrightarrow{n}}_1)+d_3({\overrightarrow{n}}_1\times {\overrightarrow{n}}_2)}{{\overrightarrow{n}}_1\cdot ({\overrightarrow{n}}_2\times {\overrightarrow{n}}_3)}.}$

For the proof one should establish${\displaystyle {\overrightarrow{n}}_i\cdot {\overrightarrow{p}}_0=d_i,i=1,2,3,}$ using the rules of ascalar triple product. If the scalar triple product equals to 0, then planes either do not have the triple intersection or it is a line (or a plane, if all three planes are the same).

Analogously to the plane case the following cases lead to non-linear systems, which can be solved using a 1- or 3-dimensional Newton iteration.^[4]

• parametric curve${\displaystyle C:(x(t),y(t),z(t))}$ and

parametric surface${\displaystyle S:(x(u,v),y(u,v),z(u,v)),}$

• parametric curve${\displaystyle C:(x(t),y(t),z(t))}$ and

implicit surface${\displaystyle S:f(x,y,z)=0.}$

Example:

parametric curve${\displaystyle C:(t,t^2,t^3)}$ and

implicit surface${\displaystyle S:x^4+y^4+z^4-1=0}$ (s. picture).

The intersection points are: (−0.8587, 0.7374, −0.6332), (0.8587, 0.7374, 0.6332).

Aline–sphere intersection is a simple special case.

Like the case of a line and a plane, the intersection of a curve and a surfaceingeneral position consists of discrete points, but a curve may be partly or totally contained in a surface.

Two transversally intersecting surfaces give anintersection curve. The most simple case is the intersection line of two non-parallel planes.

When the intersection of a sphere and a plane is not empty or a single point, it is a circle. This can be seen as follows:

LetS be a sphere with centerO,P a plane which intersectsS. DrawOE perpendicular toP and meetingP atE. LetA andB be any two different points in the intersection. ThenAOE andBOE are right triangles with a common side,OE, and hypotenusesAO andBO equal. Therefore, the remaining sidesAE andBE are equal. This proves that all points in the intersection are the same distance from the pointE in the planeP, in other words all points in the intersection lie on a circleC with centerE.^[5] This proves that the intersection ofP andS is contained inC. Note thatOE is the axis of the circle.

Now consider a pointD of the circleC. SinceC lies inP, so doesD. On the other hand, the trianglesAOE andDOE are right triangles with a common side,OE, and legsEA andED equal. Therefore, the hypotenusesAO andDO are equal, and equal to the radius ofS, so thatD lies inS. This proves thatC is contained in the intersection ofP andS.

As a corollary, on a sphere there is exactly one circle that can be drawn through three given points.^[6]

The proof can be extended to show that the points on a circle are all a common angular distance from one of its poles.^[7]

Compare alsoconic sections, which can produceovals.

To show that a non-trivial intersection of two spheres is a circle, assume (without loss of generality) that one sphere (with radius${\displaystyle R}$) is centered at the origin. Points on this sphere satisfy

${\displaystyle x^2+y^2+z^2=R^2.}$

Also without loss of generality, assume that the second sphere, with radius${\displaystyle r}$, is centered at a point on the positive x-axis, at distance${\displaystyle a}$ from the origin. Its points satisfy

${\displaystyle (x-a{)}^2+y^2+z^2=r^2.}$

The intersection of the spheres is the set of points satisfying both equations. Subtracting the equations gives

In the singular case${\displaystyle a=0}$, the spheres are concentric. There are two possibilities: if${\displaystyle R=r}$, the spheres coincide, and the intersection is the entire sphere; if${\displaystyle R\ne r}$, the spheres are disjoint and the intersection is empty.Whena is nonzero, the intersection lies in a vertical plane with this x-coordinate, which may intersect both of the spheres, be tangent to both spheres, or external to both spheres.The result follows from the previous proof for sphere-plane intersections.

• Line-plane intersection
• Line–sphere intersection
• Line-cylinder intersection
• Analytic geometry#Intersections
• Computational geometry
• Equation of a line
• Intersection (set theory)
• Intersection theory

• Hobbs, C.A. (1921).Solid Geometry. G.H. Kent. pp. 397 ff.

• Haines, Eric (June 6, 2021)."Intersections (Ray Tracing Resources Page)".Real-Time Rendering. RetrievedDecember 14, 2023.a grid of intersection routines for various popular objects, pointing to resources in books and on the web.
• Nicholas M. Patrikalakis and Takashi Maekawa,Shape Interrogation for Computer Aided Design and Manufacturing, Springer, 2002,ISBN 3540424547, 9783540424543, pp. 408.[1]
• Sykes, M.; Comstock, C.E. (1922).Solid Geometry. Rand McNally. pp. 81 ff.

• Geometric intersection

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• Short description is different from Wikidata