# Plotting cicle using ListPlot

How can I plot circle plot with ListPlot function or how to organize data to produce circle plot from ListPlot function ?

If u do something like this

ListPlot[Tuples[{{0, 1, 2, 3, 4, 5}, {-3, 3}}]]


i get

{{0, -3}, {0, 3}, {1, -3}, {1, 3}, {2, -3}, {2, 3}, {3, -3}, {3,
3}, {4, -3}, {4, 3}, {5, -3}, {5, 3}}


function value in this example is constant, and we ll see two straight lines.

but, when i need to plot circle, i have two y points per x and they are changing,

so i want plot list plot, i must have data like this:

{{0, f1[0]}, {0, f2[0]}, {1, f1[1]}, {1, f2[1]}, {2, f1[2]}, {2, f2[2]}, {3, f1[3]}, {3,f2[3]}, {4, f1[4]}, {4, f2[4]}, {5, f1[5]}, {5, f2[5]}}


where f1 is y1 and f2 is y2 and

yn = {-Sqrt[R - x^2], Sqrt[R - x^2]}


or, i can make data like this

in = Function[x, {x, (-Sqrt[25 - #1^2]) &[x], (Sqrt[25 - #1^2]) &[x]}] /@
Range[-5, 5, 1/2]
out = {{-5, 0, 0}, {-(9/2), -(Sqrt[19]/2), Sqrt[19]/2},.....


where we have {x1,y1,y2}, but List plot dosent work with this data

So, How can i plot this?

• Are you aware of ListPolarPlot? If so, why do you want to use ListPlot. Also, instead of Pythagoras, you can easily generate x and y from the Cos and Sin of the angle. Commented Apr 22, 2016 at 11:34
• p.s. CirclePoints.
– Kuba
Commented Apr 22, 2016 at 12:13
• Following up on @Kuba's comment, try this: ListPlot[CirclePoints[100], AspectRatio -> Automatic]. It's a pretty compact way of doing what you ask. Notice the AspectRatio directive in order to produce an undistorted circle. Commented Apr 22, 2016 at 14:48
• Building up on the comments above this one: Manipulate[ ListPlot[CirclePoints[i] , AspectRatio -> Automatic , PlotRange -> {{-1, 1}, {-1, 1}} , ColorFunction -> (ColorData["Rainbow"][Rescale[i, {0, 100}]] &) , Epilog -> {Text[Style[ ColorData["Rainbow"][N@Rescale[i, {1, 100}]], 20], {0.5, 0.5}]} ] , {{i, 20}, 1, 100, 1} ]
– Syed
Commented Mar 8 at 8:11

Try this:

    ListPlot[Table[{Cos[a], Sin[a]}, {a, 0, 2 \[Pi], 0.05}],
AspectRatio -> 1]


with the effect of

Have fun!

• Thanks, am soo bad at Table, can u tell me how to vary 'a' in ListPointPlot3D[ Table[{5 Cos[fi], 5 Sin[fi], a}, {fi, 0, 2 Pi, Pi/180}]] to produce cylindrical surface Commented Apr 22, 2016 at 12:17
• @Gelios Table takes more than one iterator as well. Try this for example: ListPointPlot3D[Flatten[Table[{5 Cos[fi], 5 Sin[fi], a}, {fi, 0, 2 Pi, Pi/180}, {a, -1, 1, 0.1}], 1]], which returns this plot. Commented Apr 22, 2016 at 14:46
• I see, it works, but i concerned more about data representation, i'm trying threat coordinates line a сartesian product of set(str. line and circle produces cylindrical surf.). And If i use your answer i obtain just work solution with ListPointPlot3D, but if i try this  MatrixForm@ Table[{5 Cos[fi], 5 Sin[fi], a}, {fi, 0, 2 Pi, Pi/180}, {a, -1, 1, 0.1}] it dosent make clear sence, whereas  MatrixForm@Table[{5 Cos[fi], 5 Sin[fi], a}, {fi, 0, 2 Pi, Pi/180}]  is pretty good, but 'a' should changing after one cycle and i can achieve this using AppenTo and For, but It does not look good Commented Apr 22, 2016 at 15:45
• excuse me, all ok with your solution. Commented Apr 22, 2016 at 18:55
• @Gelios MatrixForm is for visualization only, not for further calculation or plotting. Commented Apr 23, 2016 at 9:13

I guess, it will give you the circle: ListPlot[Table[{R Cos[fi],R Sin[fi]},{fi,0,2Pi,Pi/180}],AspectRatio->1]