MathGolf, 5 4 bytes

τ/╠ü

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Explanation

τ      Push tau (2*pi)

 /     Divide the first argument (total distance) by tau

  ╠    Reverse divide (computes (distance/tau)/radius)

   ü   Ceiling

APL+WIN, 9 bytes

Prompts for radius followed by distance:

⌈⎕÷○r+r←⎕

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Explanation:

○r+r←⎕ prompt for radius and double it and multiply by pie



⌈⎕÷ prompt for distance, divide by result above and take ceiling

Python 2, 47 45 44 43 bytes

lambda l,r:l/(r+r)//math.pi+l/l

import math

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• -2 bytes, thanks to flawr


• -1 byte, thanks to Jonathan Allan

Java 8, 32 30 bytes

a->b->-~(int)(a/b/Math.PI/'')

Contains unprintable u0002 between the single quotes.

Port of @jOKing's Perl 6 answer.

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Perl 6, 15 12 bytes

-3 bytes tjanks to nwellnhof reminding me about tau

*/*/τ+|$+!$

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Anonymous Whatever lambda that uses the formula (a/b/tau).floor+1. Tau is two times pi. The two anonymous variables $ are coerced to the number 0, which is used to floor the number +|0 (bitwise or 0) and add one +!$ (plus not zero).

05AB1E, 6 bytes

·/žq/î

Port of @flawr's Python 2 comment.

Takes the input in the order radius,distance.

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Explanation:

·         # Double the first (implicit) input

 /        # Divide the second (implicit) input by it

  žq/     # Divide it by PI

     î    # Ceil it (and output implicitly)

C, 46 bytes

f(float a,float b){return ceil(a/(b+b)/M_PI);}

I'm new to PPCG, so I'm not sure wether I have to count other parts in the byte count, such as the

include <math.h>

needed for the ceil function, which will rise the count to 64 bytes

Catholicon, 8 bytes

ċ//ĊǓĊ`Ė

Explanation:

  /ĊǓĊ    divide the first input by the doubled second input

 /    `Ė  divide that by pi

ċ         ceil

New version (pi builtin made one byte, division parameters swapped), 5 bytes

ċ/π/Ǔ

MathGolf, 6 5 bytes

∞/π/ü

Semi-port of @flawr's Python 2 comment.

Takes the input in the order radius distance.

-1 byte because ceil builtin has just been added, replacing the floor+1.

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Explanation:

∞        # Double the first (implicit) input

 /       # Divide the second (implicit) input by it

  π/     # Divide it by PI

    ü    # Ceil (and output implicitly)

PHP, 47 bytes

<?=ceil($argv[++$i]/M_PI/(($b=end($argv))+$b));

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Jelly, 6 bytes

÷÷ØPHĊ

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Ruby, 29 bytes

->l,r{(l/Math::PI/r+=r).ceil}

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J, 10 9 bytes

>.@%o.@+:

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JavaScript (Babel Node), 25 bytes

-2 bytes using @flawr comment =D. -1 from @Kevin. -7 from @Shaggy

a=>b=>-~(a/(b+b)/Math.PI)

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Julia 1.0, 20 bytes

f(d,r)=cld(d/π,r+r)

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R, 39 32 bytes

-7 bytes Thanks to Giuseppe

function(d,r)ceiling(d/(r+r)/pi)

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I feel like this could definitely be golfed, but I am a bit lazy right now to do anything about it

Lua, 61 58 57 bytes

s=io.read()r=io.read()print(math.ceil((s/(r+r))/math.pi))

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The inputs must be separated by a newline. I'm looking at how to read an array in LUA.
edit: Several ways, but nothing that saves bytes....

Japt, 7 bytes

/MT/V c

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Stax, 5 bytes

Vt*/e

Run and debug it

Vt*   multiply by tau (2pi)

/     divide

e     ceiling

C (gcc), 45 47 45 bytes

f(d,r,R)float d,r;{R=ceil(d/r/'G'/'n'*'q');}

A reasonable approximation of pi is 355/113. Since circumference C = 2 * r * PI, we can instead of pi use tau, which is then of course ~710/113. 710 happens to have the convenient factors 2 * 5 * 71, which is compactly expressed as 'G' * 'n'. We add one (r/r) to force rounding to infinity.

Edit: My trick was too clever for its own good: it of course made it fail if the distance was a multiple of the circumference.

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