How do I calculate intersection of two circles in polar coordinates?

I gather we're talking 2 dimensions here.

The equation of circle 2 will be (Cartesian)

x^2 + y^2 = (d1)^2

The equation of circle 1 will be (Cartesian)

(x - d1(cos o1))^2 + (y - d1(sin o1))^2 = (r1)^2

To solve these simultaneously will be very VERY messy,

so I suggest that you first calculate

h = d1(cos 01) and k = d1(sin o1),

the Cartesian coordinates of the center of circle 1.

Now you have two equations:

x^2 + y^2 = (d1)^2

(x-h)^2 + (y-k)^2 = (r1)^2

Noting that y = sqrt[ (d1)^2 - x^2 ], rewrite the 2nd equation:

(x-h)^2 + (sqrt[(d1)^2 - x^2] - k)^2 = (r1)^2

Expand the squares:

x^2 - 2xh + h^2 + (d1^2-x^2) + k^2 - 2k sqrt[(d1)^2 - x^2] = (r1)^2

Noting that h^2 + k^2 = (d1)^2, we have

2(d1)^2 - 2xh - 2k sqrt[(d1)^2 - x^2] = (r1)^2

The radical must be isolated.

2(d1)^2 - 2xh - (r1)^2 = 2k sqrt[ (d1)^2 - x^2 ]

Now square both sides:

4(d1)^4 + 4x^2 h^2 + (r1)^4 - 4 (r1)^2(d1)^2 + 4xh (r1)^2 - 8xh (d1)^2 = 4k^2 d1^2 - 4k^2 x^2

This equation is quadratic in "x" - all other symbols are constants.

However, noting again that k^2+h^2 = (d1)^2, I can write

4 x^2 (d1)^2 + x [4h (r1)^2 - 8h (d1)^2 ] + (r1)^4 - 4(r1)^2 (d1)^2 + 4(d1)^4 - 4k^2 (d1)^2 = 0.

Also I notice a perfect square near the end of this string:

4 x^2 (d1)^2 + x [4h (r1)^2 - 8h(d1)^2] + [(r1)^2 - 2(d1)^2]^2 - 4k^2 (d1)^2 = 0.

You can now use the quadratic formula:

"a" = 4 (d1)^2

"b" = 4h [ (r1)^2 - 2 (d1)^2 ]

"c" = [ (r1)^2 - 2 (d1)^2 ]^2 - 4k^2 (d1)^2 = 0.

x = the usual... = (-b plus or minus sqrt(b^2 - 4ac)/(2a).

Then go back to x^2 + y^2 = (r1)^2 to find the associated y-values;

there could be some difficulty about which of the two square roots to use,

since there will be only TWO intersection points,

and we have generated FOUR possible points.

If you are programming this,

have the computer check which two of the four points actually lie on circle 1.

Finally, "f1" is arctan(y/x), again you have a problem deciding on the quadrant,

it depends on the sign of y and the sign of x.

"f2" is obvious, it's the same as d1.

My algebraic approach is guaranteed to be correct,

but I don't guarantee perfect execution, so you should check through

the solution to make sure I didn't mess up signs or something.

This description is ambiguous - the centers of the circles must be arranged in (basically approximately*) any rhombus, and the lined section would variety. it will additionally matter on the radius, of direction. If the centers of the circles make a sq., then enable's assume circles with centers (0,0), (0,a million), (a million,0), and (a million,a million). if so, the 4-pointed shape with curved barriers would have the 4 "corners": (0.5, sqrt(3)/2), (0.5, a million-sqrt(3)/2), (sqrt(3)/2, 0.5), and (a million-sqrt(3)/2, 0.5). we are able to verify the section by using observing this as a sq. with 4 factors decrease by using chords of a circle linked to it. the part of this sq. is (sqrt(3) - a million)^2 / 2 = 2 - sqrt(3) each and every chord subtends a 30 degree ingredient of a circle (the attitude to (0.5, sqrt(3)/2) is 60 levels). the section subtended by using the chord is the part of the worldwide minus the part of a triangle. hence, the section over the chord would be: A(sector) = Pi (a million^2) /12 = Pi/12 A(triangle) = a million/2(a million) (a million) (sin 30) = 0.25, so A(over chord) = Pi/12 - 0.25 A(4 chords) = Pi/3 - a million So the total part of this shape for a unit circle is: A = 2 - sqrt(3) + Pi/3 - a million = a million + Pi/3 - sqrt(3) ~= 0.315 (and of direction, for a regular length circle, you will multiply this by using r^2.) * A rhombus with attitude 60 levels does not count huge form - 2 of the circles would have circumferences that flow by way of ALL 3 of the others. additionally, if the circles may be distinctive sizes the form may be a kite with one end-attitude of 60 levels.

The polar form of a circle always results in the two equations of the quadratic formula:

r = {-b + √{b² - 4(c)}/2 and r = {-b - √{b² - 4(c)}/2

(Notice that I left out the coefficient for the square term (a); this is, because it must always be 1 for a circle)

I will derive the above for you:

(x - h)² + (y - k)² = r1²

where (h, k) is the center point and r1 is the radius

To convert to polar coordinates, let h = {d1cos(o1)}, k = {d1sin(01)}, x = (r)cos(θ), and y = (r)sin(θ)

((r)cos(θ) - {d1cos(o1)})² + ((r)sin(θ) - {d1sin(o1)})² = r1²

r²cos²(θ) - 2{d1cos(o1)}(r)cos(θ) + {d1cos(o1)}² + r²sin²(θ) - 2{d1sin(o1)}(r)sin(θ) + {d1sin(o1)}² = r1²

Using cos²(x) + sin²(x) = 1:

r² - 2d1{(cos(o1)cos(θ) + sin(o1)sin(θ)}r + d1² - r1² = 0

This a = 1, b = -2d1{(cos(o1)cos(θ) + sin(o1)sin(θ)}, and c = d1² - r1²

r = {2d1{(cos(o1)cos(θ) + sin(o1)sin(θ)} + √[-2d1{(cos(o1)cos(θ) + sin(o1)sin(θ)}² - 4d1² + 4r1²]}/2

and

r = {2d1{(cos(o1)cos(θ) + sin(o1)sin(θ)} - √[-2d1{(cos(o1)cos(θ) + sin(o1)sin(θ)}² - 4d1² + 4r1²]}/2

Because the second circle is at the origin, its equation is r = r2, therefore, you can set the two equation equal to r2:

r2 = {2d1{(cos(o1)cos(θ) + sin(o1)sin(θ)} + √[-2d1{(cos(o1)cos(θ) + sin(o1)sin(θ)}² - 4d1² + 4r1²]}/2

and

r2 = {2d1{(cos(o1)cos(θ) + sin(o1)sin(θ)} - √[-2d1{(cos(o1)cos(θ) + sin(o1)sin(θ)}² - 4d1² + 4r1²]}/2

When you get values for the constants, this will be easy to solve for the two values of θ.