Add region constraint to GraphicsHow to add fill color to a 2D Graphics object?How to use Graphics@Rotate@Show@Graphics without “Graphics is not a Graphics primitive or directive.”How to embed a filled Region in a Graphics?Integrated Solution GraphicsEdgeForm with GraphicsCombine graphics with GraphicsGridHow can I add engineering-style dimensioning to graphics?Is it possible to overlay pixel graphics with vector graphicsHow to add graphics lines in a grid?How can I get high precision circle drawing in Graphics, across ten orders of magnitude

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Add region constraint to Graphics


How to add fill color to a 2D Graphics object?How to use Graphics@Rotate@Show@Graphics without “Graphics is not a Graphics primitive or directive.”How to embed a filled Region in a Graphics?Integrated Solution GraphicsEdgeForm with GraphicsCombine graphics with GraphicsGridHow can I add engineering-style dimensioning to graphics?Is it possible to overlay pixel graphics with vector graphicsHow to add graphics lines in a grid?How can I get high precision circle drawing in Graphics, across ten orders of magnitude






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








5












$begingroup$


I would like to only show the part of Black circles within the Red circle.



Can I do that with Graphics, Circle and/or some sort of Region Constraint?



the upper half



step = 2 Degree;
[Alpha] = Range[2 Degree, 80 Degree, step];
x = ConstantArray[0, Length@[Alpha]];
y = Sec@[Alpha];
radius = Tan@[Alpha];
range = 1.01;
Graphics[Red, Circle[0, 0, 1], Black,
MapThread[Circle[#1, #2, #3] &, x, y, radius],
PlotRange -> -range, range, -range, range]


At the moment, I use ContourPlot with RegionFunction option. However, my plot contains a large number of these circles (The amount of the circles shown here is only a quarter for the current step), which makes ContourPlot approach very slow. Moreover, when I zoom in, quite often I find the circles drawn by ContourPlot are not circular, presumably due to PlotPoints and MaxRecursion etc. I tried to play with these two options, but did not succeed in terms of quality (being a circle) and speed.



Thank you!










share|improve this question









$endgroup$


















    5












    $begingroup$


    I would like to only show the part of Black circles within the Red circle.



    Can I do that with Graphics, Circle and/or some sort of Region Constraint?



    the upper half



    step = 2 Degree;
    [Alpha] = Range[2 Degree, 80 Degree, step];
    x = ConstantArray[0, Length@[Alpha]];
    y = Sec@[Alpha];
    radius = Tan@[Alpha];
    range = 1.01;
    Graphics[Red, Circle[0, 0, 1], Black,
    MapThread[Circle[#1, #2, #3] &, x, y, radius],
    PlotRange -> -range, range, -range, range]


    At the moment, I use ContourPlot with RegionFunction option. However, my plot contains a large number of these circles (The amount of the circles shown here is only a quarter for the current step), which makes ContourPlot approach very slow. Moreover, when I zoom in, quite often I find the circles drawn by ContourPlot are not circular, presumably due to PlotPoints and MaxRecursion etc. I tried to play with these two options, but did not succeed in terms of quality (being a circle) and speed.



    Thank you!










    share|improve this question









    $endgroup$














      5












      5








      5





      $begingroup$


      I would like to only show the part of Black circles within the Red circle.



      Can I do that with Graphics, Circle and/or some sort of Region Constraint?



      the upper half



      step = 2 Degree;
      [Alpha] = Range[2 Degree, 80 Degree, step];
      x = ConstantArray[0, Length@[Alpha]];
      y = Sec@[Alpha];
      radius = Tan@[Alpha];
      range = 1.01;
      Graphics[Red, Circle[0, 0, 1], Black,
      MapThread[Circle[#1, #2, #3] &, x, y, radius],
      PlotRange -> -range, range, -range, range]


      At the moment, I use ContourPlot with RegionFunction option. However, my plot contains a large number of these circles (The amount of the circles shown here is only a quarter for the current step), which makes ContourPlot approach very slow. Moreover, when I zoom in, quite often I find the circles drawn by ContourPlot are not circular, presumably due to PlotPoints and MaxRecursion etc. I tried to play with these two options, but did not succeed in terms of quality (being a circle) and speed.



      Thank you!










      share|improve this question









      $endgroup$




      I would like to only show the part of Black circles within the Red circle.



      Can I do that with Graphics, Circle and/or some sort of Region Constraint?



      the upper half



      step = 2 Degree;
      [Alpha] = Range[2 Degree, 80 Degree, step];
      x = ConstantArray[0, Length@[Alpha]];
      y = Sec@[Alpha];
      radius = Tan@[Alpha];
      range = 1.01;
      Graphics[Red, Circle[0, 0, 1], Black,
      MapThread[Circle[#1, #2, #3] &, x, y, radius],
      PlotRange -> -range, range, -range, range]


      At the moment, I use ContourPlot with RegionFunction option. However, my plot contains a large number of these circles (The amount of the circles shown here is only a quarter for the current step), which makes ContourPlot approach very slow. Moreover, when I zoom in, quite often I find the circles drawn by ContourPlot are not circular, presumably due to PlotPoints and MaxRecursion etc. I tried to play with these two options, but did not succeed in terms of quality (being a circle) and speed.



      Thank you!







      graphics






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Jul 14 at 3:36









      Bemtevi77Bemtevi77

      1457 bronze badges




      1457 bronze badges




















          2 Answers
          2






          active

          oldest

          votes


















          6












          $begingroup$

          You can use the three-argument form of Circle:



          Graphics[Red, Circle[0, 0, 1], Black, 
          MapThread[Circle[#1, #2, #3, π + ArcTan[#3], 2 π - ArcTan[#3]] &,
          x, y, radius],
          PlotRange -> -range, range, -range, range]


          enter image description here



          Alternatively, use RegionIntersection with Disk[] to get the needed portions of black circles:



          circles = MapThread[Circle[#1, #2, #3] &, x, y, radius];

          circles2 = RegionIntersection[Disk[], #] & /@ N[circles];

          Graphics[Red, Circle[0, 0, 1], Black, circles2,
          PlotRange -> -range, range, -range, range]



          same picture




          Update: An alternative way to hide unwanted portions of circles using FilledCurve:



          filledCurve = FilledCurve[Line[Append[#, First @ #]& @ 
          CirclePoints[range Sqrt @2, 4]],
          Line[Append[#, First @ #]& @ CirclePoints[200]]];

          Graphics[Red, Circle[0, 0, 1], Black, circles,
          EdgeForm[None], White, filledCurve,
          PlotRange -> -range, range, -range, range]



          same picture as above







          share|improve this answer











          $endgroup$












          • $begingroup$
            I used the 1st method of yours, which is the fastest among all the provided answers at the moment. But it seems the ArcTan[#3 Sqrt[1/(1 + #3^2)] Sqrt[1 + #3^2]] can be shortened to ArcTan[#3]. Also, the center of the circles lies along the y axis, which is a special case. I wonder what this method could be if the center of the circle is arbitrary. Could you please elaborate if possible? Thanks. I realized that for the "arbitrary center" case, it requires the actual information of the circle, and therefore, it maybe changes from case to case.
            $endgroup$
            – Bemtevi77
            Jul 14 at 6:33











          • $begingroup$
            @Bemtevi77, updated with the simpler form. Re arbitrary centers, I think the second and third methods should work as is. I think the first method should also work but I am not sure.
            $endgroup$
            – kglr
            Jul 14 at 6:42


















          5












          $begingroup$

          g = Graphics[Red, Circle[0, 0, 1], Black, 
          MapThread[Circle[#1, #2, #3] &, x, y, radius],
          PlotRange -> -range, range, -range, range];

          Show[g, RegionPlot[x^2 + y^2 > 1, x, -1.2, 1.2, y, -1.2, 1.2,
          PlotStyle -> White]]


          enter image description here






          share|improve this answer









          $endgroup$












          • $begingroup$
            I realized that this method can be easily applied to "circles with arbitrary center and radius" as asked in my comment to kglr's answer. It may not be the fastest, but is more versatile, in my opinion. Thank you very much!
            $endgroup$
            – Bemtevi77
            Jul 14 at 6:43













          Your Answer








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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          6












          $begingroup$

          You can use the three-argument form of Circle:



          Graphics[Red, Circle[0, 0, 1], Black, 
          MapThread[Circle[#1, #2, #3, π + ArcTan[#3], 2 π - ArcTan[#3]] &,
          x, y, radius],
          PlotRange -> -range, range, -range, range]


          enter image description here



          Alternatively, use RegionIntersection with Disk[] to get the needed portions of black circles:



          circles = MapThread[Circle[#1, #2, #3] &, x, y, radius];

          circles2 = RegionIntersection[Disk[], #] & /@ N[circles];

          Graphics[Red, Circle[0, 0, 1], Black, circles2,
          PlotRange -> -range, range, -range, range]



          same picture




          Update: An alternative way to hide unwanted portions of circles using FilledCurve:



          filledCurve = FilledCurve[Line[Append[#, First @ #]& @ 
          CirclePoints[range Sqrt @2, 4]],
          Line[Append[#, First @ #]& @ CirclePoints[200]]];

          Graphics[Red, Circle[0, 0, 1], Black, circles,
          EdgeForm[None], White, filledCurve,
          PlotRange -> -range, range, -range, range]



          same picture as above







          share|improve this answer











          $endgroup$












          • $begingroup$
            I used the 1st method of yours, which is the fastest among all the provided answers at the moment. But it seems the ArcTan[#3 Sqrt[1/(1 + #3^2)] Sqrt[1 + #3^2]] can be shortened to ArcTan[#3]. Also, the center of the circles lies along the y axis, which is a special case. I wonder what this method could be if the center of the circle is arbitrary. Could you please elaborate if possible? Thanks. I realized that for the "arbitrary center" case, it requires the actual information of the circle, and therefore, it maybe changes from case to case.
            $endgroup$
            – Bemtevi77
            Jul 14 at 6:33











          • $begingroup$
            @Bemtevi77, updated with the simpler form. Re arbitrary centers, I think the second and third methods should work as is. I think the first method should also work but I am not sure.
            $endgroup$
            – kglr
            Jul 14 at 6:42















          6












          $begingroup$

          You can use the three-argument form of Circle:



          Graphics[Red, Circle[0, 0, 1], Black, 
          MapThread[Circle[#1, #2, #3, π + ArcTan[#3], 2 π - ArcTan[#3]] &,
          x, y, radius],
          PlotRange -> -range, range, -range, range]


          enter image description here



          Alternatively, use RegionIntersection with Disk[] to get the needed portions of black circles:



          circles = MapThread[Circle[#1, #2, #3] &, x, y, radius];

          circles2 = RegionIntersection[Disk[], #] & /@ N[circles];

          Graphics[Red, Circle[0, 0, 1], Black, circles2,
          PlotRange -> -range, range, -range, range]



          same picture




          Update: An alternative way to hide unwanted portions of circles using FilledCurve:



          filledCurve = FilledCurve[Line[Append[#, First @ #]& @ 
          CirclePoints[range Sqrt @2, 4]],
          Line[Append[#, First @ #]& @ CirclePoints[200]]];

          Graphics[Red, Circle[0, 0, 1], Black, circles,
          EdgeForm[None], White, filledCurve,
          PlotRange -> -range, range, -range, range]



          same picture as above







          share|improve this answer











          $endgroup$












          • $begingroup$
            I used the 1st method of yours, which is the fastest among all the provided answers at the moment. But it seems the ArcTan[#3 Sqrt[1/(1 + #3^2)] Sqrt[1 + #3^2]] can be shortened to ArcTan[#3]. Also, the center of the circles lies along the y axis, which is a special case. I wonder what this method could be if the center of the circle is arbitrary. Could you please elaborate if possible? Thanks. I realized that for the "arbitrary center" case, it requires the actual information of the circle, and therefore, it maybe changes from case to case.
            $endgroup$
            – Bemtevi77
            Jul 14 at 6:33











          • $begingroup$
            @Bemtevi77, updated with the simpler form. Re arbitrary centers, I think the second and third methods should work as is. I think the first method should also work but I am not sure.
            $endgroup$
            – kglr
            Jul 14 at 6:42













          6












          6








          6





          $begingroup$

          You can use the three-argument form of Circle:



          Graphics[Red, Circle[0, 0, 1], Black, 
          MapThread[Circle[#1, #2, #3, π + ArcTan[#3], 2 π - ArcTan[#3]] &,
          x, y, radius],
          PlotRange -> -range, range, -range, range]


          enter image description here



          Alternatively, use RegionIntersection with Disk[] to get the needed portions of black circles:



          circles = MapThread[Circle[#1, #2, #3] &, x, y, radius];

          circles2 = RegionIntersection[Disk[], #] & /@ N[circles];

          Graphics[Red, Circle[0, 0, 1], Black, circles2,
          PlotRange -> -range, range, -range, range]



          same picture




          Update: An alternative way to hide unwanted portions of circles using FilledCurve:



          filledCurve = FilledCurve[Line[Append[#, First @ #]& @ 
          CirclePoints[range Sqrt @2, 4]],
          Line[Append[#, First @ #]& @ CirclePoints[200]]];

          Graphics[Red, Circle[0, 0, 1], Black, circles,
          EdgeForm[None], White, filledCurve,
          PlotRange -> -range, range, -range, range]



          same picture as above







          share|improve this answer











          $endgroup$



          You can use the three-argument form of Circle:



          Graphics[Red, Circle[0, 0, 1], Black, 
          MapThread[Circle[#1, #2, #3, π + ArcTan[#3], 2 π - ArcTan[#3]] &,
          x, y, radius],
          PlotRange -> -range, range, -range, range]


          enter image description here



          Alternatively, use RegionIntersection with Disk[] to get the needed portions of black circles:



          circles = MapThread[Circle[#1, #2, #3] &, x, y, radius];

          circles2 = RegionIntersection[Disk[], #] & /@ N[circles];

          Graphics[Red, Circle[0, 0, 1], Black, circles2,
          PlotRange -> -range, range, -range, range]



          same picture




          Update: An alternative way to hide unwanted portions of circles using FilledCurve:



          filledCurve = FilledCurve[Line[Append[#, First @ #]& @ 
          CirclePoints[range Sqrt @2, 4]],
          Line[Append[#, First @ #]& @ CirclePoints[200]]];

          Graphics[Red, Circle[0, 0, 1], Black, circles,
          EdgeForm[None], White, filledCurve,
          PlotRange -> -range, range, -range, range]



          same picture as above








          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Jul 14 at 6:59

























          answered Jul 14 at 5:07









          kglrkglr

          207k10 gold badges236 silver badges468 bronze badges




          207k10 gold badges236 silver badges468 bronze badges











          • $begingroup$
            I used the 1st method of yours, which is the fastest among all the provided answers at the moment. But it seems the ArcTan[#3 Sqrt[1/(1 + #3^2)] Sqrt[1 + #3^2]] can be shortened to ArcTan[#3]. Also, the center of the circles lies along the y axis, which is a special case. I wonder what this method could be if the center of the circle is arbitrary. Could you please elaborate if possible? Thanks. I realized that for the "arbitrary center" case, it requires the actual information of the circle, and therefore, it maybe changes from case to case.
            $endgroup$
            – Bemtevi77
            Jul 14 at 6:33











          • $begingroup$
            @Bemtevi77, updated with the simpler form. Re arbitrary centers, I think the second and third methods should work as is. I think the first method should also work but I am not sure.
            $endgroup$
            – kglr
            Jul 14 at 6:42
















          • $begingroup$
            I used the 1st method of yours, which is the fastest among all the provided answers at the moment. But it seems the ArcTan[#3 Sqrt[1/(1 + #3^2)] Sqrt[1 + #3^2]] can be shortened to ArcTan[#3]. Also, the center of the circles lies along the y axis, which is a special case. I wonder what this method could be if the center of the circle is arbitrary. Could you please elaborate if possible? Thanks. I realized that for the "arbitrary center" case, it requires the actual information of the circle, and therefore, it maybe changes from case to case.
            $endgroup$
            – Bemtevi77
            Jul 14 at 6:33











          • $begingroup$
            @Bemtevi77, updated with the simpler form. Re arbitrary centers, I think the second and third methods should work as is. I think the first method should also work but I am not sure.
            $endgroup$
            – kglr
            Jul 14 at 6:42















          $begingroup$
          I used the 1st method of yours, which is the fastest among all the provided answers at the moment. But it seems the ArcTan[#3 Sqrt[1/(1 + #3^2)] Sqrt[1 + #3^2]] can be shortened to ArcTan[#3]. Also, the center of the circles lies along the y axis, which is a special case. I wonder what this method could be if the center of the circle is arbitrary. Could you please elaborate if possible? Thanks. I realized that for the "arbitrary center" case, it requires the actual information of the circle, and therefore, it maybe changes from case to case.
          $endgroup$
          – Bemtevi77
          Jul 14 at 6:33





          $begingroup$
          I used the 1st method of yours, which is the fastest among all the provided answers at the moment. But it seems the ArcTan[#3 Sqrt[1/(1 + #3^2)] Sqrt[1 + #3^2]] can be shortened to ArcTan[#3]. Also, the center of the circles lies along the y axis, which is a special case. I wonder what this method could be if the center of the circle is arbitrary. Could you please elaborate if possible? Thanks. I realized that for the "arbitrary center" case, it requires the actual information of the circle, and therefore, it maybe changes from case to case.
          $endgroup$
          – Bemtevi77
          Jul 14 at 6:33













          $begingroup$
          @Bemtevi77, updated with the simpler form. Re arbitrary centers, I think the second and third methods should work as is. I think the first method should also work but I am not sure.
          $endgroup$
          – kglr
          Jul 14 at 6:42




          $begingroup$
          @Bemtevi77, updated with the simpler form. Re arbitrary centers, I think the second and third methods should work as is. I think the first method should also work but I am not sure.
          $endgroup$
          – kglr
          Jul 14 at 6:42













          5












          $begingroup$

          g = Graphics[Red, Circle[0, 0, 1], Black, 
          MapThread[Circle[#1, #2, #3] &, x, y, radius],
          PlotRange -> -range, range, -range, range];

          Show[g, RegionPlot[x^2 + y^2 > 1, x, -1.2, 1.2, y, -1.2, 1.2,
          PlotStyle -> White]]


          enter image description here






          share|improve this answer









          $endgroup$












          • $begingroup$
            I realized that this method can be easily applied to "circles with arbitrary center and radius" as asked in my comment to kglr's answer. It may not be the fastest, but is more versatile, in my opinion. Thank you very much!
            $endgroup$
            – Bemtevi77
            Jul 14 at 6:43















          5












          $begingroup$

          g = Graphics[Red, Circle[0, 0, 1], Black, 
          MapThread[Circle[#1, #2, #3] &, x, y, radius],
          PlotRange -> -range, range, -range, range];

          Show[g, RegionPlot[x^2 + y^2 > 1, x, -1.2, 1.2, y, -1.2, 1.2,
          PlotStyle -> White]]


          enter image description here






          share|improve this answer









          $endgroup$












          • $begingroup$
            I realized that this method can be easily applied to "circles with arbitrary center and radius" as asked in my comment to kglr's answer. It may not be the fastest, but is more versatile, in my opinion. Thank you very much!
            $endgroup$
            – Bemtevi77
            Jul 14 at 6:43













          5












          5








          5





          $begingroup$

          g = Graphics[Red, Circle[0, 0, 1], Black, 
          MapThread[Circle[#1, #2, #3] &, x, y, radius],
          PlotRange -> -range, range, -range, range];

          Show[g, RegionPlot[x^2 + y^2 > 1, x, -1.2, 1.2, y, -1.2, 1.2,
          PlotStyle -> White]]


          enter image description here






          share|improve this answer









          $endgroup$



          g = Graphics[Red, Circle[0, 0, 1], Black, 
          MapThread[Circle[#1, #2, #3] &, x, y, radius],
          PlotRange -> -range, range, -range, range];

          Show[g, RegionPlot[x^2 + y^2 > 1, x, -1.2, 1.2, y, -1.2, 1.2,
          PlotStyle -> White]]


          enter image description here







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Jul 14 at 4:57









          MelaGoMelaGo

          2,1161 gold badge1 silver badge7 bronze badges




          2,1161 gold badge1 silver badge7 bronze badges











          • $begingroup$
            I realized that this method can be easily applied to "circles with arbitrary center and radius" as asked in my comment to kglr's answer. It may not be the fastest, but is more versatile, in my opinion. Thank you very much!
            $endgroup$
            – Bemtevi77
            Jul 14 at 6:43
















          • $begingroup$
            I realized that this method can be easily applied to "circles with arbitrary center and radius" as asked in my comment to kglr's answer. It may not be the fastest, but is more versatile, in my opinion. Thank you very much!
            $endgroup$
            – Bemtevi77
            Jul 14 at 6:43















          $begingroup$
          I realized that this method can be easily applied to "circles with arbitrary center and radius" as asked in my comment to kglr's answer. It may not be the fastest, but is more versatile, in my opinion. Thank you very much!
          $endgroup$
          – Bemtevi77
          Jul 14 at 6:43




          $begingroup$
          I realized that this method can be easily applied to "circles with arbitrary center and radius" as asked in my comment to kglr's answer. It may not be the fastest, but is more versatile, in my opinion. Thank you very much!
          $endgroup$
          – Bemtevi77
          Jul 14 at 6:43

















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