Plotting two circles and finding their intersection points

Question

I would like to generate the two circle graphic shown below:

As the first step, I want to make the first circle with center (0,0) and diameter of 100 mm by using the program that follows:

But, I feel confused when I will make second circle with center of distance 85 mm and diameter of 140 mm.

My questions:

1. Is there another better technique to make the first circle and the second circle? Also, please advice me how to get points in both circles?

2. How to find the intersection point between the first circle and the second circle?

Answer 1

rg1 = Circle[{0, 0}, 2];
rg2 = Circle[{0., 1}, 1.5];
pts=RegionIntersection[rg1, rg2] 
RegionPlot[{rg1, rg2}, AspectRatio -> Automatic, 
 Epilog -> {PointSize -> Large, pts}]

Point[{{-1.45237, 1.375}, {1.45237, 1.375}}]

Answer 2

1.

You can use the Circle Expression to create your circles instead of plotting points. Then poke the circle expressions into Graphics to create the plot. (Since you want axes, we set the Axis option to True)

circle1 = Circle[{0, 0}, 50];
circle2 = Circle[{85, 0}, 70];
Graphics[{circle1, circle2}, Axes -> True]

2.

To find the intersection of the circle we can use the formulas for the two circles and the solve function

equation1 = x^2 + y^2 == 50^2;
equation2 = (x - 85)^2 + y^2 == 70^2;
solutions = Solve[{equation1, equation2}]

To convert that into a nice set of coordinates, we use Part which can be written as [[ ]]. It is similar to indexing in other programming languages.

coords = solutions[[ All, All, 2]]

Then we generate points using Point and Map (which can also be written as /@) to generate Point expressions to insert into the Graphics function. Map (when written as /@) wraps every item of the top level expression on the right hand side with the function supplied on the left hand side.

points = Point /@ coords

The we join our points with the circles Join to feed to the Graphics function similar to what we did in part 1

Graphics[Join[{circle1, circle2}, points], Axes -> True]

To make the points easier to see, we can make them larger with the PointSize Graphics directive

Graphics[Join[{PointSize[Large], circle1, circle2}, points], Axes -> True]

Answer 3

Just for illustrative purposes:

i[{c1_, r1_}, {c2_, r2_}] := 
 Module[{r}, 
  If[c1 == c2 && r1 == r2, {}, 
   r = {x, y} /. 
     Solve[{({x, y} - c1).({x, y} - c1) == 
        r1^2, ({x, y} - c2).({x, y} - c2) == r2^2}, {x, y}, Reals]; 
   If[r === {x, y}, {}, r]]]
f[{c1_, r1_}, {c2_, r2_}, lim_] :=
 Module[{sol = Quiet@i[{c1, r1}, {c2, r2}]}, 
  Graphics[{Circle[c1, r1], Circle[c2, r2], Red, PointSize[0.02], 
    Point[sol], Text[c1, c1], Text[c2, c2], Blue,
    Arrow[{c1, c1 + r1 {-1/Sqrt[2], 1/Sqrt[2]}}], 
    Text[Framed[r1, Background -> White], 
     c1 + r1 {-1/Sqrt[2], 1/Sqrt[2]}/2],
    Arrow[{c2, c2 + r2 {1/Sqrt[2], 1/Sqrt[2]}}], 
    Text[Framed[r2, Background -> White], 
     c2 + r2 {1/Sqrt[2], 1/Sqrt[2]}/2],
    Text[#, #, {-1.5, 0}] & /@ sol
    }, PlotRange -> ConstantArray[{-lim, lim}, 2], Axes -> True, 
   GridLines -> Transpose[{c1, c2}]]]
f[{{0, 0}, 50}, {{85, 0}, 70}, 170]
Manipulate[
 f[{c1, r1}, {c2, r2}, 2], {c1, {-1, -1}, {1, 1}, 
  Locator}, {{c2, {0, 0}}, {-1, -1}, {1, 1}, Locator}, {r1, 0.1, 
  2}, {{r2, 1}, 0.1, 2}]

Answer 4

I would like to add the alternative answer for you.

C1 = Table[{50*Cos[\[Theta]*Degree], 
 50*Sin[\[Theta]*Degree]}, {\[Theta], 1, 360}] // N;
C2 = Table[{70*Cos[\[Theta]*Degree] + 85, 
 70*Sin[\[Theta]*Degree]}, {\[Theta], 1, 360}] // N;

To show in graphic, you can use:

ListPlot[{C1, C2}, AspectRatio -> Automatic]

For the intersection, you can follow program below:

equation1 = x^2 + y^2 == 50^2;
equation2 = (x - 85)^2 + y^2 == 70^2;
solutionMax = Solve[{equation1, equation2 && y > 0}];
solutionMin = Solve[{equation1, equation2 && y < 0}];

coordMax = solutionMax[[All, All, 2]]
coordMin = solutionMin[[All, All, 2]]
{{965/34, (35 Sqrt[1599])/34}}
{{965/34, -((35 Sqrt[1599])/34)}}

Answer 5

It is straightforward to do it with Region functionality. However, if equations are the starting point, then the following could be a possibility.

Clear["Global`*"]
eq1 = x^2 + y^2 - 50^2;
eq2 = (x - 85)^2 + y^2 - 70^2;

etyp = (-h + x)^2 + (-k + y)^2 == r^2;

sol1 = SolveAlways[ eq1 == Subtract @@ etyp, {x, y}];
sol2 = SolveAlways[ eq2 == Subtract @@ etyp, {x, y}];

c1 = Circle[{h, k}, r] /. First@Pick[sol1, r /. allsol // UnitStep, 1]
c2 = Circle[{h, k}, r] /. First@Pick[sol2, r /. allsol // UnitStep, 1]

Circle[{0, 0}, 50]

Circle[{85, 0}, 70]

ContourPlot[{eq1, eq2}
 , {x, -60, 160}, {y, -80, 80}
 , MeshStyle -> Directive[PointSize[Large], Red]
 , Mesh -> {{0}}
 , MeshFunctions -> {Function[{x, y}, eq1], Function[{x, y}, eq2]}
 , AspectRatio -> Automatic
 , Axes -> True
 , AxesStyle -> {{Gray, DotDashed}, {Gray, DotDashed}}
 , Epilog -> {
   Black, AbsolutePointSize[6]
   , Point@{RegionCentroid@c1
     , RegionCentroid@c2}
   
   , {Dashed, Arrowheads[{-.05, .05}]
    , Arrow[CirclePoints[{Last@c1, π/2 - π/14}, 2]]}
   
   , {Dashed, Arrowheads[{-.05, .05}]
    , GeometricTransformation[
     Arrow[CirclePoints[{Last@c2, π/3}, 2], 0.01]
     , TranslationTransform[First@c2]
     ] (* end of GeometricTransformation *)
    }
   } (* end of Epilog *)
 ]