# Intersection of ellipse with circle

I would like know whether a circle is intersecting an ellipse.

Here ellipse equation is $$Ax^2 + Bxy + Cy^2 + dx+ey + 1 = 0,$$

and the circle equation is

$$(x-g)^2 + (y-f)^2= r^2.$$

• You have to solve the equations for real numbers to see if they intersect. Notice that depending on your choice of $a$ to $g$, you can have intersection in $0$ to $4$ points. – BlackAdder Jan 29 '14 at 7:22

The circle can be written as $$x=g + r \frac{1-t^2}{1+t^2} \quad ; \quad y= f + r \frac{2t}{1+t^2}.$$ Substitute these expressions for $x$ and $y$ into your ellipse equation. This will give a polynomial equation of degree four in $t$. The real solutions (if any) give the intersection points.

Another approach is to construct a pair of lines that pass through the points of intersection. Look up the concept of "pencils" of conics to find out more about this approach. For example, the technique is discussed on this page: http://en.m.wikipedia.org/wiki/Conic_section

You can try to create functions based on you ellipse and circle equation and then build a system out of them. To do that, isolate one of the variables (e.g., y) in both equations and equate y for the circle and y for the ellipse.

For example, functions for the circle centered at the origin and with radius one can obtained like this $$x^2+y^2=1 \implies y^2=1-x^2\implies y=\pm \sqrt{1-x^2}$$

You can treat the circle as a special kind of ellipse. Then you are in the domain of Finding Intersection of an ellipse with another ellipse when both are rotated, and my answer there applies to your case. You still have to solve cubic equations for that approach, which by the way is this “pencil of conics” stuff bubba mentions in his answer. In the last step of my answer, one possible approach is intersecting a line with a conic, which might be simpler if you intersect that line with your circle, not with the other ellipse.

Rotate the ellipse by $\tan^{-1}\frac{ B} { A-C }$ to bring them to form where axes are parallel to axes.

To get a feel, plotting them parametrically will help.The ellipse plots with

$$x = H + A \cos t ,\,y= K + B \sin t ;$$

Next plot circles by varying $h,k,a$

$$x = h + a \cos u, \, y=k + a \sin u$$

along with ellipse to see if and where they cut.

When $t,u$ are both real in the following equation then cutting takes place.

$$K + B \sin t = k + a \sin u$$

When one of them is imaginary then they do not cut. It two real roots are coincident or repeated then they touch.