# 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? – Dingus Sep 22 '20 at 6:49
• yes, it doesn't make sense. I changed it. – Razetime Sep 22 '20 at 6:53
• Related (Draw a Polygon) – Jo King Sep 22 '20 at 8:04
• You say there should be a padding of 5 units, what should happen if radius is bigger than window size? – LiefdeWen Sep 22 '20 at 8:36
• If we can zoom, wouldn't the same constant-sized circle, not showing axes, work for any input? – att Sep 23 '20 at 6:58

# 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 – Giuseppe Sep 22 '20 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. – Therkel Sep 23 '20 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...? – Dominic van Essen Sep 23 '20 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. – Giuseppe Sep 23 '20 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 :-) – Giuseppe Sep 23 '20 at 15:03

# 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.

# 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. – Razetime Sep 23 '20 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) . – Kjetil S. Sep 24 '20 at 21:57

# 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 '20 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. – Razetime Sep 22 '20 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 – Digital Trauma Sep 22 '20 at 16:26
• This is a really cool approach, and I can't believe there aren't more turtle golfs in Python! – Dominic van Essen Sep 23 '20 at 8:06
• @DominicvanEssen I came up with turtle solution as a first result in Googling "graphics python". – Arty Sep 23 '20 at 9:09

# SageMath, 24 bytes

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


Try it online!

### Example # 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: # 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? – Razetime Oct 28 '20 at 11:02
• Oh, I missed that part. Let me change it. – corvus_192 Oct 28 '20 at 11:11

# Desmos, 13 bytes

x^2+y^2=r^2
r


$$x^2 + y^2 = r^2 \\{} r$$

Desmos it • Is there a way to center the plot/zoom in desmos using code? – Razetime Sep 22 '20 at 6:15
• @Razetime if you've zoomed away from centre, press the home button. – lyxal Sep 22 '20 at 6:16
• I mean, for r=50, the circle extends outside the screen. is there a way to zoom to fit? – Razetime Sep 22 '20 at 6:17
• i mean, programatically. – Razetime Sep 22 '20 at 6:26
• MATL should work for that. It's available online. – Razetime Sep 22 '20 at 6:54

## 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. – Razetime Sep 22 '20 at 9:46 # 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. # 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. – Greg Martin Sep 22 '20 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. – ZaMoC Sep 22 '20 at 19:10 • You could just use Graphics@Circle@##&, or shorter yet, Graphics@*Circle. – LegionMammal978 Sep 24 '20 at 12:16 • @LegionMammal978 yes, you're right.. – ZaMoC Sep 24 '20 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. – Tom Carpenter Sep 24 '20 at 17:49 # 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 – Aaron Miller Jun 7 at 14:12 • @AaronMiller Thank you! – Galen Ivanov Jun 7 at 14:16 ## <>^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) : # 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. – Razetime Sep 23 '20 at 6:35 • Yes @Razetime thanks, I tried but it's really hard from a phone :/ – AZTECCO Sep 23 '20 at 6:41 • oh, gotcha. I think border-radius 50% is enough, so -1 byte. – Razetime Sep 23 '20 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? – Dominic van Essen Sep 23 '20 at 7:59 • The circle is always at the center of the screen, due to top and left:50% – Razetime Sep 23 '20 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);} – PrincePolka Jul 6 at 15:51 • My above code might not apply with the challenge, does print a circle though – PrincePolka Jul 8 at 18:00 # 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! – Galen Ivanov Jun 2 at 8:39 # TI-Basic, 11 bytes Input R ZSquare Circle(0,0,R  (TI-basic is tokenized) Output for R=4: • +1, looks like sniping target XD – wasif Jun 14 at 7:42 # 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


# PHP, 131 bytes

($f=imagecolorallocate)($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. – Razetime Oct 1 '20 at 10:16

# 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\$$: # Python + pygame, 110 bytes

r=int(input())
d=display
s=5+r
draw.circle(d.set_mode([s+s]*2),*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.

# 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
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]
; 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
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: # HTML + Javascript, 97

## JavaScript: 84, HTML: 13

r=100;C.width=C.height=r*2+16;c=C.getContext2d;c.arc(r+8,r+8,r,0,6.283);c.stroke()
<canvas id=C>

To change r, change r=100 to r=/*place your value here*/.