# Intro

Given radius $$\r\$$, draw a circle in the center of the screen.

Sandbox.

# The Challenge

Here is a simple challenge.

Plot a circle using the formula $$\x^2+y^2=r^2\$$, or any other formula that will plot a circle according to the given parameters.

You may use any units that your language provides, so long as they are well defined and give consistent output.

The circle must have it's center at the center of the canvas, and must have a padding of 5 units or more on all sides.

The circle can have any fill that does not match the outline.

You may have axes in the background of your plot.

The outline of the circle must be solid (no gaps), and it must be visible. Here is an example:

Input can be taken in any acceptable form. (function params, variables, stdin...)

Output can be in the form of a separate window, or an image format.

Standard loopholes and rules apply.

# Example Code (Java + Processing)

// Modified from the C language example from
// https:// en.wikipedia.org/wiki/Midpoint_circle_algorithm
void settings() {
size(2*r+10, 2*r+10);
}

void draw() {
background(255);
drawCircle(width/2, height/2, r, 60);
save("Circle.png");
}

void drawCircle(int x0, int y0, int radius, int angle) {
int circCol = color(0, 0, 0);
int y = 0;
int err = 0;

while (x >= y && atan2(y, x) < limit) {
set(x0 + x, y0 + y, circCol);
set(x0 + y, y0 + x, circCol);
set(x0 - y, y0 + x, circCol);
set(x0 - x, y0 + y, circCol);
set(x0 - x, y0 - y, circCol);
set(x0 - y, y0 - x, circCol);
set(x0 + y, y0 - x, circCol);
set(x0 + x, y0 - y, circCol);

y += 1;
if (err <= 0) {
err += 2*y + 1;
}
if (err > 0) {
x -= 1;
err -= 2*x + 1;
}
}
}


# Scoring

This is a question. No ascii art.

This is . shortest answer in each language wins.

• What do you mean by 'the thickness of the outline must be 1 unit'? Are the units of outline thickness the same as those of the radius? Sep 22, 2020 at 6:49
• yes, it doesn't make sense. I changed it. Sep 22, 2020 at 6:53
• Related (Draw a Polygon)
– Jo King
Sep 22, 2020 at 8:04
• You say there should be a padding of 5 units, what should happen if radius is bigger than window size? Sep 22, 2020 at 8:36
• If we can zoom, wouldn't the same constant-sized circle, not showing axes, work for any input?
– att
Sep 23, 2020 at 6:58

# Desmos, 1 byte

r

Try it on Desmos

Uses the same input method as the other Desmos answer. An unused variable named r defaults to drawing a circle with radius r.

# R, 74706865 54 bytes

Edit: -11 bytes thanks to Giuseppe

function(r)plot(r*1i^(1:1e3/99),,"l",l<-c(-r-5,r+5),l)


Try it at rdrr.io

I propose 3 possible answers to this challenge in R, of decreasing length but with increasing caveats.

My favourite answer (above, using plot) is the middle shortest one of the 3. It plots a circle by calculating the complex coordinates of powers of i, using 396 points (with a bit of wrap-around). Here's an image of the output from plot_circle(5):

For a 'true' circle (rather than an almost-circle with tiny straight-lines connecting the data points), we can use the curve function with a formula, but unfortunately we need to draw the positive & negative halves separately, so it ends up longer:

# R, 86 84 bytes

function(r){curve((r^2-x^2)^.5,xli=l<-c(-r-5,r+5),yli=l)


Try it at rdrr.io

The shortest (that I can think of) Previously the shortest, but - thanks to Giuseppe now no longer so - is to use the circles option of the symbols function, for only 56 bytes. However, this has the caveat that the circle symbols are always circular even if the plot is re-sized, and so may no-longer line-up with the y-axis.

# R, 6258 56 bytes

function(r)symbols(x=1,c=r,i=F,xli=l<-c(-r-4,r+6),yli=l)


Try it at rdrr.io

• would pty="s" work instead of setting the xlim and ylim for the first 2? I was digging through the documentation for par but haven't had a chance to test it. It says "s" generates a square plotting region Sep 22, 2020 at 16:18
• @Giuseppe par(pty="s") should work but then it wouldn't satisfy the condition of 5 units of padding. In fact, from what I can see, only this answer satisfies the padding condition! Without par(pty="s") the circle depends on the output device being square, since the device will otherwise stretch. Sep 23, 2020 at 7:15
• @Therkel - welcome (back) to code-golf, and, in case you weren't aware, R is the language of the month this month (Sept 2020)! As you obviously know R, how about giving a shot at a couple of R-golfs...? Sep 23, 2020 at 7:37
• @Therkel I completely missed the 5 padding requirement. Seems a bit random to be honest. At the least, you can use the order of variables in plot to get the first down to 54 bytes; presumably similar golfs would work for the others. Sep 23, 2020 at 14:24
• well, I googled "R plot" and went to the first result from the stat.ethz.ch site (which I always do when googling R documentation), which luckily was the page for plot.default! The docs for plot say plot.default will be used for simple scatterplots, so I suppose I lucked out somewhat :-) Sep 23, 2020 at 15:03

# Python 2 & 3, 55 bytes

-17 bytes thanks to @DigitalTrauma
-1 byte thanks to @Sisyphus
-2 bytes thanks to @ovs

from turtle import*
def f(r):sety(-r);clear();circle(r)


Try it online!

turtle is standard library included in Python 2 & 3. I came up with turtle idea as a almost first result in Googling "graphics python".

• If window is allowed to be closed automatically after drawing is done, then done() can be removed from end of code, it just prevents window from closing.
– Arty
Sep 22, 2020 at 16:10
• so long as it displays the circle, even for a frame, there is no problem. I think you can remove int() from int(input()) as well. Sep 22, 2020 at 16:11
• fd() is an alias of forward(). Also you can use goto() to set the start position: 60 bytes with the other suggestions Sep 22, 2020 at 16:26
• This is a really cool approach, and I can't believe there aren't more turtle golfs in Python! Sep 23, 2020 at 8:06
• @DominicvanEssen I came up with turtle solution as a first result in Googling "graphics python".
– Arty
Sep 23, 2020 at 9:09

# SageMath, 24 bytes

lambda r:circle((0,0),r)


Try it online!

# LaTeX, 66 bytes

\input tikz\def\f#1{~\vfill\centering\tikz\draw circle(#1);\vfill}


I consider the "canvas" required by this challenge to be the default text area of a latex page. The code defines a macro \f that takes the radius in cm as an argument.

## Example Code

\documentclass{article}

\input tikz\def\f#1{~\vfill\centering\tikz\draw circle(#1);\vfill}

\begin{document}
\f{3}
\enddocument


Outputs a PDF:

• How is the input given? Oct 28, 2020 at 11:02
• Oh, I missed that part. Let me change it. Oct 28, 2020 at 11:11

# C (gcc), 169166161 160 bytes

Thanks to ceilingcat (x3) for the suggestions! I also changed the newlines in the header to spaces as they seem to work fine as separators (at least in Irfanview) and fixed a bug that got revealed when the array was put on the stack.

Generates an image in PBM format, as it's probably the simplest way to make a bitmap! For some reason, all the online PBM viewers that I tried don't seem to like the output file, but Irfanview and GIMP are fine with it.

z;f(r,w){char s[(w=r*2+11)*w+1];float x=s[w*w]=!memset(s,48,w*w);for(;x<7;)s[z=round(sin(x+=1e-5)*r+r+5)+round(cos(x)*r+r+5)*w]=49;printf("P1 %d %d %s",w,w,s);}


Try it online!

• Libreoffice shows the correct circle image. Nice Answer. Sep 23, 2020 at 3:38
• Save 21 bytes with z,x;f(r,w){char s[(w=r*2+11)*w+1];for(x=!memset(s,48,w*w);x<7e5;)s[z=sin(++x)*r+r+5+round(cos(x)*r+r+5)*w]=49;printf("P1 %d %d %s",w,w,s);}. No need for the first round and to keep x below 7 (or 2π) . Sep 24, 2020 at 21:57

# T-SQL, 47 bytes

SELECT geometry::Point(0,0,0).STBuffer(r)FROM t


Input is taken via a pre-existing table t with float field r, per our IO rules.

Uses SQL geo-spatial functions, displayed in the SQL Management Studio results pane:

## JavaScript (ES6), 96 bytes

f=
v=><svg width=${s=v*2+12} height=${s}><circle r=${v} cx=${v+=6} cy=${v} stroke=#000 fill=none> <input type=number min=1 oninput=o.innerHTML=f(+this.value)><div id=o> Outputs an SVG(HTML5) image, which the snippet displays for you. If HTML5 is acceptable, then for 95 bytes: f= v=><div style="width:${v*=2}px;height:${v}px;margin:6px;border:1px solid;border-radius:100%"> <input type=number min=1 oninput=o.innerHTML=f(+this.value)><div id=o> • feel free to use html5 for your main answer. Sep 22, 2020 at 9:46 # Wolfram Language (Mathematica), 16 bytes Graphics@*Circle  for example here is a circle with center (0,0) and radius r=42 -6 bytes from @LegionMammal978 • The specifications call for a single input, the radius. On the other hand, Mathematica should center the circle in the plot window automatically, regardless of the center. Sep 22, 2020 at 15:40 • @GregMartin "...or any other formula that will plot a circle according to the given parameters". Do you thing you you have a shortest solution? I don't think that Circle[] is acceptable, so this is the shortest way. Sep 22, 2020 at 19:10 • You could just use Graphics@Circle@##&, or shorter yet, Graphics@*Circle. Sep 24, 2020 at 12:16 • @LegionMammal978 yes, you're right.. Sep 24, 2020 at 12:35 # MATLAB, 38 37 bytes -1 byte thanks to Tom Carpenter ezpolar(@(x)r);axis((r+5)*cospi(1:4))  Input as variable r in workspace. Output: • Ignore my previous deleted comments, forgot the circle had to be centred. You can save a byte with ezpolar(@(x)r);axis((r+5)*cospi(1:4)). The cospi(1:4) gives you [-1 1 -1 1] which can be multipled directly with (r+5) to get your limits array. Sep 24, 2020 at 17:49 ## <>^v, 56 bytes ƒ∆57±∑361i∆90v v 0Ii(I< ª < ] ^ >¶° ^ >1 æ∑ ^  #### Explanation ƒ : toggle turtle visibility ∆ : raise pen 57: push 57 ± : negate ∑ : turtle forward by top of stack 361: push 361 (360 + 1) i : pop stack & store in variable i ∆ : lower pen 90: push 90 v : send instruction pointer down < : send instruction pointer left ª : turtle rotate right by top of stack (90) start of loop < : send instruction pointer left I : push value of variable i ( : decrement top of stack i : pop & store in variable i I : push value of variable i 0 : push 0 v : send instruction pointer down ] : if top of stack is greater than or equal to second element of stack\ • > : send instruction pointer right • ¶ : update display • continue to continue here\ Else > : send instruction pointer right 1: push 1 æ : turn left (top of stack) degrees ∑ : go forward by top of stack ^ : send instruction pointer up continue here ^ : send instruction pointer up ^ : Idem, there to ensure trailing whitespace is not removed go to start of loop After drawing the circle, the program never halts to prevent the window from closing. Screenshot below (Python Turtle Graphics is because the program does not set a title to the window and the interpreter is written in Python and uses Turtle for graphics) : • Welcome to Code Golf! Jun 13, 2021 at 23:57 • does this take input? Mar 18 at 4:55 • @Razetime No, it doesn't Mar 18 at 11:53 • Just realized how stupid what I just said was. Didn't see the program ought to take a radius input. I'll update it as soon as possible. Mar 18 at 11:56 # Desmos, 12 10 bytes xx+yy=rr r  Desmos it • Is there a way to center the plot/zoom in desmos using code? Sep 22, 2020 at 6:15 • @Razetime if you've zoomed away from centre, press the home button. Sep 22, 2020 at 6:16 • I mean, for r=50, the circle extends outside the screen. is there a way to zoom to fit? Sep 22, 2020 at 6:17 • i mean, programatically. Sep 22, 2020 at 6:26 • MATL should work for that. It's available online. Sep 22, 2020 at 6:54 # TI-Basic, 11 8 7 bytes Polar Input r₁ ZSquare  Input is a string, and is stored as a function. This function is shown in polar coordinates, so it shows as a circle. Output for "4: • +1, looks like sniping target XD Jun 14, 2021 at 7:42 # Java 8, 141 123 bytes import java.awt.*;r->new Frame(){{setSize(2*r+26,2*r+56);show();}public void paint(Graphics g){g.drawOval(13,43,2*r,2*r);}}  Output for $$\n=100\$$ (the second picture with added light grey background color is to verify the top padding): Explanation: import java.awt.*; // Required import for Frame and Graphics r-> // Method with integer parameter and Frame return-type new Frame(){ // Create the Frame { // In an inner code-block: setSize(2*r // Set the width to 2 times the radius-input +26 // + 2 times 8 pixels for the frame borders // + 2 times 5 pixels for the required padding 2*r // Set the height to 2 times the radius-input +56); // + 2 times 8 pixels for the frame borders // + 30 pixels for the frame title-bar // + 2 times 5 pixels for the required padding show();} // And show the Frame at the end public void paint(Graphics g){ // Overwrite its paint method to draw on g.drawOval(13,43, // With 5,5 for the required padding as top-left // x,y-coordinate of the surrounding rectangle + the same 8+30 // pixels adjustment for the frame and frame title-bar, 2*r,2*r);}} // draw the circle with a size of 2 times the radius-input  Note: I cannot use (r*=2),r,r,r instead of 2*r,2*r,2*r,2*r to save (3) bytes, because r has to be effectively final inside the inner Frame-class. # Red, 74 bytes -16 bytes thanbks to Aaron Miller! func[r][d: r + 5 view compose/deep[base(2x2 * d)draw[circle(1x1 * d)(r)]]]  # Red, 90 bytes func[r][d: r + 5 view compose/deep[base(as-pair d * 2 d * 2)draw[circle(as-pair d d)(r)]]]  f 200 • 74 bytes Jun 7, 2021 at 14:12 • @AaronMiller Thank you! Jun 7, 2021 at 14:16 # Python + pygame, 110 bytes r=int(input()) d=display s=5+r draw.circle(d.set_mode([s+s]*2),[99]*3,[s]*2,r,1) d.update()  We only use set_mode once, so we can pass it as an argument to draw.circle. It's almost impossible to golf in Pygame so this is probably the shortest it can get, although I could change 99 to 9. Example for 250: # JavaScript (V8), 158 bytes r=>document.write(<p style="border-radius:50%;border:solid;position:fixed;width:${r*2}px;height:${r*2}px;top:50%;left:50%;transform:translate(-50%,-50%);">)  Try it online! jsfiddle thanks to @Razetime • Saved 1 thanks to @Razetime • Saved 2 using template literals Writes directly to the HTML a p element fixed positioned, centered with border radius 50% • This is better hosted on jsfiddle. Sep 23, 2020 at 6:35 • Yes @Razetime thanks, I tried but it's really hard from a phone :/ Sep 23, 2020 at 6:41 • oh, gotcha. I think border-radius 50% is enough, so -1 byte. Sep 23, 2020 at 6:43 • The jsfiddle link just gives me a bouncing cloud with an infinity sign... So (without having actually witnessed the output): does border-radius 50% satisfy the "must have a padding of 5 units or more on all sides" when r<10? Sep 23, 2020 at 7:59 • The circle is always at the center of the screen, due to top and left:50% Sep 23, 2020 at 15:17 ## Shadertoy (GLSL), 142 bytes void mainImage(out vec4 f,in vec2 v){vec2 S=iResolution.xy;vec2 u=v/S-vec2(0.5);u.y/=S.x/S.y;vec4 c;if(abs(length(u)-0.2)<8e-4)c=vec4(1);f=c;}  Shadertoy link Output: • 85 bytes: void mainImage(out vec4 f,in vec2 v){v-=vec2(400,225);f=vec4(abs(dot(v,v)-1e4)<1e2);} Jul 6, 2021 at 15:51 • My above code might not apply with the challenge, does print a circle though Jul 8, 2021 at 18:00 # Red, 9692 87 bytes Forgot to remove some extra whitespace for -4 bytes. -5 bytes by using shorter type conversions and initializing and using x at the same time. draw to-pair x: r * 9 to-block append"circle "append mold to-pair x / 2 append" "mold r  No TIO link because draw doesn't seem to be implemented on TIO. However, you can copy this into the Red offline interpreter to output an image. The first line is for defining a variable to be used for the canvas size. Multiplying by 9 might be a bit overkill, but it ensures enough padding around the circle. I couldn't figure out how to use variables in the block, so the second line builds the draw command bit by bit, essentially building the command draw {x}x{x} [circle {x / 2}x{x / 2} {r}]. Example output for $$\r = 10\$$: # Red, 595755 51 byte func[r][?(draw 2 * c: 5x5 + r reduce['circle c r])]  Try it locally. -2 bytes thanks to Aaron Miller spotting superfluous as-pair. Output for $$\r=50\$$: • Nice solution! Great use of the tools provided by the GUI console! Jun 2, 2021 at 8:39 # 80186 DOS machine code, 767472 71 bytes 00000000: b8 13 00 cd 10 1e 68 00 a0 1f f7 df 8d 6d 02 01 ......h......m.. 00000010: fd 31 c0 f7 df e8 20 00 f7 df 97 e8 1a 00 89 eb .1.... ......... 00000020: 39 df e8 0a 00 97 39 fb e8 04 00 75 e6 1f 7c 06 9.....9....u..|. 00000030: 01 fd 8d 6b 03 47 c3 e8 00 00 69 d8 40 01 fe 81 ...k.G....i.@... 00000040: a0 7d f7 df f7 d8 c3 .}.....  A function which expects the radius in di. Forces DOS to graphics mode and doesn't bother to clean up. I am mostly just excited about getting it working, I will hopefully find some ways to optimize. Translation of the Go version of the Midpoint circle algorithm from Rosetta Code. • 2 bytes: branch once in correct instead of branching twice in the main function • 2 bytes: reuse flags from inc for loop trigger. • 1 byte: Fall through to correct when returning. #### Commented assembly  ; nasm file.asm -f obj -o file.obj [cpu 186] global draw_circle ; args: di: radius draw_circle: ; switch DOS to graphics mode mov ax, 0x0013 int 0x10 ; set ds to point to the screen buffer push ds push 0xA000 pop ds ; http://rosettacode.org/wiki/Bitmap/Midpoint_circle_algorithm#Go ; di => x1 neg di ; bp => err lea bp, [2 + di] add bp, di ; ax => y1 xor ax, ax .loop: neg di ; x - x1, y + y1 ; x + x1, y - y1 call set_pixel neg di ; swap xchg ax, di ; x + y1, y + x1 ; x - y1, y - x1 call set_pixel ; ax: x1, di: y1 ; save err to bx mov bx, bp ; check y1 ; we branch in correct cmp di, bx call correct ; swap back xchg ax, di ; check x1, using the opposite order cmp bx, di call correct ; loop while x1 is negative ; the flags will be set from correct jnz .loop ; restore ds segment pop ds ; uncomment to wait for enter then switch to ; standard mode ; mov ah, 0x08 ; int 0x21 ; mov ax, 0x0003 ; int 0x10 ; Fallthrough to exit ; some helper functions to cut down the copy-paste correct: ; The flags will be set to the comparison before ; calling jl .skip ; err += 2 * ++di + 1 add bp, di lea bp, [bp + di + 3] ; ++di inc di .skip: ret ; ax: y, di: x set_pixel: ; run twice by semi-recursion call .semirecurse .semirecurse: ; ++byte[x + 160 + (320 * (y + 100))] imul bx, ax, 320 inc byte[di + bx + 0x7DA0] ; negate both here neg di neg ax ret  with radius = 50, showing the disgusting exit: ## ZX Spectrum BASIC, 1, 3 or 6 bytes Nothing beats BASIC computers with bitmap graphics that consider a case of honour to implement the built in CIRCLE command... CIRCLE x,y,r The command takes just one byte (though it is printed as the whole word), depending on what and how do you count the size, it can be 1 byte (the command itself), 3 bytes (the command and obligatory commas separating the articles; note that the space is only visual), 6 bytes (the command, commas and three already defined variables (or PI)). • you must input radius and center it in the screen. x and y are not specified inputs, and must not be taken. predefined variables are not allowed in a submission. Mar 19 at 1:40 • @Razetime from the question: "Input can be taken in any acceptable form. (function params, variables, stdin...)" - and these are function (or procedure, if you prefer) parameters. Also, the original example does not use any inputs, but defines variables. I think some clarification is needed... Mar 19 at 5:50 • Ok, sorry, only one predefined variable is allowed. r is the only allowed input. you must define your entire program to make sure that it does not use any input other than r (and center the circle). Mar 19 at 8:08 # Desmos, 10 12 bytes xx+yy=rr r=1  Desmos it # Perl 5, 92 bytes $r=$_;$w=$r*2+11;$_="P1 $w$w @{[map{($_%$w-$r-5)**2+($_/$w-$r-5)**2<$r**2?1:0}0..$w**2-1]}"


Try it online!

Circle with black filling. Put the 92 bytes above into program.pl and run like this:

echo 50 | perl -p program.pl > circle.pbm  # radius 50
feh circle.pbm                             # view with feh or other image viewer


($f=imagecolorallocate)($i=imagecreate($d=10+$argn*2,$d),0,0,0);imageellipse($i,$d/2,$d/2,$d-10,$d-10,$f($i,9,9,9));imagepng($i,a);  Try it online! Actually you cannot run it in online PHP testers because they disable the image functions. Saves the image in a file named "a". One byte could be saved using imagegd but I didn't know the "gd" format and couldn't open it to check if it works. The circle is in very dark grey, but I consider it visible. If you don't, leave a comment and I'll edit, with one byte more $f($i,99,0,0) it's much clearer. with $f($i,9,9,9): with $f($i,99,0,0): • dark grey is different from black, so it's alright. I can see the circle. Oct 1, 2020 at 10:16 # Swift, entire iOS app, 152 bytes project.pbxproj not included. import SwiftUI @main struct A:App{@State var s="1" var body:some Scene{WindowGroup{VStack{TextField("",text:$s)
Circle().frame(width:.init(Int(s)!))}}}}


The unit for the radius is 0.5pt, which is either 1px or 1.5px depending on the device. The app crashes if you type something that isn't an integer.

Preview, with a radius of 50pt:

### Ungolfed

import SwiftUI

@main
struct MyApp: App {
var body: some Scene {
WindowGroup {
ContentView()
}
}
}

struct ContentView: View {
@State private var text = "1"

var body: some View {
VStack {
TextField("", text: $text) Circle() .frame(width: CGFloat(Int(text)!)) } } }  • does this take input? Mar 18 at 4:54 • No, it doesn’t respond to user interaction at all. Mar 18 at 11:40 • the question requires a method of input to scale the circle. Mar 18 at 11:49 • Oh, you’re right, I missed that. I can’t fix it now but I can get to it later today. Mar 18 at 11:50 ## PostScript, 69 bytes currentpagedevice/PageSize get{2 div}forall 3 2 roll 0 360 arc stroke  Pass the radius on the command line: gs -c 150 -- circle.ps  Displays output in a window, which then immediately closes! Add showpage for a persistent window. Centering the output took more bytes than I had hoped... # HTML + Javascript, 98 ## JavaScript: 85, HTML: 13 F=r=>{C.width=C.height=r*2+16;c=C.getContext2d;c.arc(r+8,r+8,r,0,6.283);c.stroke()} F(100) <canvas id=C> To change r, change r=100 to r=/*place your value here*/. • you need to provide a function or take the result of an alert() Mar 18 at 4:56 • @Razetime Fixed! May 2 at 23:08 # C (gcc), 132 128 bytes -4 bytes thanks to @ceilingcat main(r,w,i,j){scanf("%d",&r);i=w=r+9;for(printf("P5\n%d%1$d\n1\n",2*w);i--+w;)for(j=w;j--+w;)putchar(abs((i*i+j*j)*2-r*r)<2*r);}


Try it online!

Outputs a PGM image. A unit corresponds to sqrt(2) pixels.

## Applesoft BASIC, 106 bytes

0INPUTR
1HGR
2HCOLOR=3
3D=0
4HPLOTR*COS(D)+140,R*SIN(D)+96TOR*COS(D+.1)+140,R*SIN(D+.1)+96
5D=D+0.1
6GOTO4