Two ellipses intersection

Ellipse

x² +

y² +

x +

y +

= 0

(x −	)²	+	( y −	)²	= 1

2			2

x² +

y² +

x +

y +

= 0

(x −	)²	+	( y −	)²	= 1

2			2

Intersection points:

Intersection point 1:

Intersection point 2:

Intersection point 3:

Intersection point 4:

Common area

Input limit:

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If both ellipses centers are located at the origin (0 , 0) the equations of the ellipses will be:

The intersection points are at:

If the ellipses are given by the polynomial forms Ax² + By² + E = 0 and Fx² + Gy² + J = 0

Then the intersection points are given by:

There are three posible locations of the ellipses:

#	Relative position	Total intersection points	Condition
①		0 points	a > c and b > d or a < c and b < d
②		2 points	a = c and b ≠ d or b = d and a ≠ c
③		4 points	a > c and b < d or a < c and b > d

The lapping area of ellipse of type ① and ② is the area of the inner ellipse A = πab or A = πcd

The area of ellipse type ③ is:

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The equation of an ellipse which is located along the x (y = 0) axis is:

Mark:

α = a²d² − b²

β = hb² − ma²d²

γ = a²d²m² + a²b² − b²h² − a²d²c²

And the x coordinate of the intersections are:

Or after substituting values:

The polynomial form of this ellipse is Ax² + By² + Cx + E = 0 and Fx² + Gy² + Hx + J = 0

After substituting the value of y from eq. (1) to eq. (2) we get the quadratic equation with x:

x²(AG − BF) + x(CG − BH) + (EG − BJ) = 0

And the intersection is:

There are six possible locations of the ellipses:

(the value of D in this case is: D = |m − h| )

#	Relative position	Total intersection points	Condition
①		0 points	a + c < D (along x axis) or b + d < D (along y axis)
②		1 points	a + c = D or b + d = D
③		2 point	a + c > D or b + d > D
④		3 points	D + c = a or D + d = b
⑤		4 points	D + c < a and D < fy(m) or D + d < b and D < fx(n)
⑥		0 points	a + c < D and D > fy(m) or b + d < D and D > fy(n)

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If the ellipses are given by the equations

From the second equation the value of y is:

And for the value of x we get a quartic equation:

Where:

The equations of this ellipses by polynomial forms are:

Ax² + By² + Cx + Dy + E = 0	(1)
Fx² + Gy² + J = 0	(2)

From eq. (1) we get the value of y:

Substitute the value of y into equation (2) yields the equation:

After multiplying all terms by 4B² and open paranthesis we get the equation

In order to simplepy the equation we will define the following constants:

δ =	4B²J + 2D²G − 4BEG
φ =	4BCG
θ =	4B²F − 4ABG
μ =	2DG

After arranging the equation by powers we get a forthe power of x which can be solved mathematically:

x⁴θ² − x³2θφ + x²(φ² + 2θδ + 4Bμ²) − x2φδ + (δ² − D²μ² + 4BEμ²) = 0

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If the ellipses are given by the equsations:

After quite many stepses we get a quartic equation for the x coordinate of the intersection:

θx⁴ + 2θμx³ + (μ² + 2θγ + b²δ)x² + (2μγ − 2b²hδ²)x + (γ² − αδ²) = 0

Where:	α = b² (a² − h²)

The y coordinate is given by the equation:

We define two ellipses by the polynomial equations:

Ax² + By² + Cx + Dy + E = 0	(1)
Fx² + Gy² + Hx + Iy + J = 0	(2)

From eq. (1) we get the value of y:

Substitute the value of y into equation (2) yields the equation:

After multiplying all terms by 4B² and open paranthesis we get the equation

In order to simplepy the equation we will define the following constants:

δ =	2BDI − 2D²G + 4BEG − 4B²J
φ =	4BCG − 4B²H
θ =	4ABG − 4B²F
μ =	2BI − 2DG

And we get the equation:

Taking the squares of bothe sides we get:

D²μ² − 4ABμ²x² − 4BCμ²x − 4BEμ² = x⁴θ² + x²φ² + δ² + 2θφx³ + 2θδx² + 2φδx

After arranging the equation by powers we get a forthe power of x which can be solved mathematically:

x⁴θ² + x³2θφ + x²(φ² + 2θδ + 4ABμ²) + x(4BCμ² + 2φδ) + (δ² − D²μ² + 4BEμ²) = 0