Two circles intersection calculator

Print two circle intersection calculator

Circles form: (x-a)² + (y-b)² = r²
Circle 1:	( x + )² + ( y + )² = 2
Circle 2:	( x + )² + ( y + )² = 2

Circles form: x² + y² + Ax + By + C = 0
Circle 1:	x² + y² + x + y + = 0
Circle 2:	x² + y² + x + y + = 0

Intersection coordinates (x , y):
Line passing through intersection points:
Distance between circles centers:
Equation of the line passing circles centers:
Common area between circles:
Circle 1 - x and y axis intercepts:
Circle 2 - x and y axis intercepts:

Notes:

Two circles tangency summary
Intercepts of a circle and the axis

JScript code to find intersection points Example 1 Example 2 Example 3 Example 4 Example 5

Two circles intersection equations summary

Two circles tangency conditions summary

Description	Condition
	Circles equations: (x + x₁)² + (y + y₁)² = r₁² (x + x₂)² + (y + y₂)² = r₂² Distance between circles centers d is: d < \|r₁ − r₂\|
	Inner tangency d = \|r₁ − r₂\|
	Intersecting circles \|r₁ − r₂\| < d < r₁ + r₂
	Outer tangency d = r₁ + r₂
	Intersecting circles d > r₁ + r₂

Intercepts of a circle and the x and y axis summary

Circle equation: (x − a)² + (y − b)² = r²

In order to find the circle intercept with the y axis substitute the value
x = 0 in the circle equation and solve for y.

a² + (y − b)² = r²

y² − 2by + a² + b² − r² = 0

We got a quadratic equation with the y unknown.

x_1,2 = 0

When r > a then two y axis intercepts points exists.

When r = a then the circle is tangent to the y axis.

When r < a then the circle does not intercept with the y axis.

Apply the same steps to find the intercept with the x axis (where y = 0)

y_1,2 = 0

When r > b then two x axis intercepts points exists.

When r = b then the circle is tangent to the x axis.

When r < b then the circle does not intercepts with the x axis.

Circle equation: x² + y² + Ax + By + C = r²

Substitute x = 0 for intercept with the y axis: y² + By + C = 0

x_1,2 = 0

Substitute y = 0 for the intercepts with the x axis: x² + Ax + C = 0

y_1,2 = 0

Example 1 - x and y axes intercepts

Find the x and y axis intercepts points of the circle: x² + y² + 6x − 16 = 0.

For the y axis intercepts substitute x = 0 into the circle equation: y² − 16 = 0.

And the y axis intercepts are: y₁ = 4 y₂ = −4.

For the x axis intercepts substitute y = 0 into the circle equation: x² + 6x − 16 = 0.

The solution of this quadratic equation are:

And the x axis intercepts are: x₁ = 2 x₂ = −8

Example 2 - x and y axes intercepts

Find the x and y axes intercepts points of the circle (x − 4)² + (y + 1)² = 16.

for the y axis intercepts insert the value: x = 0 to get:

y² + 2y + 1 = 0

The solution of this equation is:
Because there is only one solution to this equation, the circle must be tangent to the y axis at point (0 , −1).

For the x axis intercepts we insert the value: y = 0 And get the quadratic equation: x² − 8x + 1 = 0.

The solution of this equation is:

And the x axes intercepts are at: (7.873 , 0) and (0.127 , 0).

Example 3 - Two circles overlapping area

Find the overlapping area between the two circles given by the equations:
(x + 3)² + (y + 2)² = 36 and (x − 4)² + (y + 2)² = 16.

First solution will be solved by geometric method.

The blue area is a circle segment of the left circle with cord equal to 2h so the blue area is:

A(r,θ)=r^2/2 (θ-sin⁡θ) (1)

If the two circles are: (x + a)² + (y + b)² = r₁² and (x + c)² + (y + d)² = r₂²

The distance between circles centers is:
	D = A + B	(2)
The heigh h can be found from the triangles:	h² = r₁² − A²	(3)
	h² = r₂² − B²	(4)
From equations (3) and (4) we have	r₁² − r₂² = B² − A²	(5)

Substitute A and B from equation (2) into equation (5) to get:

From triangle r₁, h, A angle #θ₁ can be found:
The same way angle θ₂ is:

The blue segment area is found according to equation (1)

And the total common area of two circles is:

A_common=A_blue+A_green=〖r_1〗^2/2 (θ_1-sin⁡〖θ_1 〗 )+〖r_2〗^2/2 (θ_2-sin⁡〖θ_2 〗 )

From the given circles the distance between circles centers is:

And the angles θ₁ and θ₂ are:

2θ_1=2 cos^(-1)⁡〖A/r_1 〗=2 cos^(-1)⁡〖4.929/6〗=1.213

2θ_2=2 cos^(-1)⁡〖B/r_2 〗=2 cos^(-1)⁡〖2.071/4〗=2.053

Another way to find the common area is by integration

From the equation of a circle, we have:
And the blue segment area is:

From the sketch we see that the boundaries of the integration are x₁ = A and x₂ = r₁

The solution is performed by trigonometric substitution method and is equal to:

A_blue=∫_(x_1)^(x_2)▒√(〖r_1〗^2-x^2 ) dx

JScript code to calculate two circles intersection points

//Definition of a circle object

function circle(a, b, r) { this.a = a; this.b = b; this.r = r; }

// This values should be declared as global variables

// so their values can be used without return their values

var x1, y1, x2, y2;

function calculateIntersection() {

// Calling function

var circle1 = new circle(2, 4, 5);

var circle2 = new circle(-1, 0, 3);

if (twoCirclesIntersection(circle1, circle2)) {

// If true - then the circles intersect

// the intersection points are given by (x1, y1) and (x2, y2)

..... Continue with desirable code .....

}

function twoCirclesIntersection(c1, c2){

//**************************************************************

//Calculating intersection coordinates (x1, y1) and (x2, y2) of

//two circles of the form (x - c1.a)^2 + (y - c1.b)^2 = c1.r^2

// (x - c2.a)^2 + (y - c2.b)^2 = c2.r^2

// Return value: true if the two circles intersect

// false if the two circles do not intersect

//**************************************************************

var val1, val2, test;

// Calculating distance between circles centers

var D = Math.sqrt((c1.a - c2.a) * (c1.a - c2.a) + (c1.b - c2.b) * (c1.b - c2.b));

if (((c1.r + c2.r) >= D) && (D >= Math.abs(c1.r - c2.r))) {

// Two circles intersects or tangent

// Area according to Heron's formula

//----------------------------------

var a1 = D + c1.r + c2.r;

var a2 = D + c1.r - c2.r;

var a3 = D - c1.r + c2.r;

var a4 = -D + c1.r + c2.r;

var area = Math.sqrt(a1 * a2 * a3 * a4) / 4;

// Calculating x axis intersection values

//---------------------------------------

val1 = (c1.a + c2.a) / 2 + (c2.a - c1.a) * (c1.r * c1.r - c2.r * c2.r) / (2 * D * D);

val2 = 2 * (c1.b - c2.b) * area / (D * D);

x1 = val1 + val2;

x2 = val1 - val2;

// Calculating y axis intersection values

//---------------------------------------

val1 = (c1.b + c2.b) / 2 + (c2.b - c1.b) * (c1.r * c1.r - c2.r * c2.r) / (2 * D * D);

val2 = 2 * (c1.a - c2.a) * area / (D * D);

y1 = val1 - val2;

y2 = val1 + val2;

// Intersection points are (x1, y1) and (x2, y2)

// Because for every x we have two values of y, and the same thing for y,

// we have to verify that the intersection points as chose are on the

// circle otherwise we have to swap between the points

test = Math.abs((x1 - c1.a) * (x1 - c1.a) + (y1 - c1.b) * (y1 - c1.b) - c1.r * c1.r);

if (test > 0.0000001) {

// point is not on the circle, swap between y1 and y2

// the value of 0.0000001 is arbitrary chose, smaller values are also OK

// do not use the value 0 because of computer rounding problems

var tmp = y1;

y1 = y2;

y2 = tmp;

}

return true;

}

else {

// circles are not intersecting each other

return false;

}

Example 4 - Two circles tangency coordinate

Find the points of intersections of the circles (x + 1)² + (y + 1)² = 4 and (x − 4)² + (y + 1)² = 9.

The term (y + 1)² is the same in the equations of both circles. From the first circle we have:

(y + 1)² = 4 − (x + 1)²

substitute this value to the first circle:

(x + 1) + 9 − (x − 4) − 4 = 0

After solving the equation we receive x_t = 1

Now we can find the value of y_t

And y_t = −1 or tangency at (1 , −1)

Example 5 - Two circles intersection

Find the common cord length of the two circles (x + h)² + (y + k)² = r² and (x + k)² + (y + h)² = r²

Find the numerical values of the cord length, when h = 3 k = -1 first circle center at (-3, 1) and r = 4

Because both radii are the same we can write

(x + h)² + (y + k)² = (x + k)² + (y + h)²

after manipulating this equation and eliminating same values we get the solution:

(h − k)(x − y) = 0

First solution is: h = k

Substituting this value in both circles equation we see that both has same center at point (−k , −k) and because the radii are the same, both circles are identical.

Second solution is when y = x and the circle equation becomes:

(x + h)² + (x + k)² = r²

2x² + 2(h + k)x + h² + k² − r² = 0

The solutions of this quadratic equation are: