Exploding Cats!

JavaScript (ES6), 211 216 220

Edit 1 byte saved thx @usandfriends

s=>[...s].map((c,i)=>{for(R=_=>Math.random()*h-l|0;x=R(y=R()),!(i-(l/2|0)?x|y&&x*x+y*y<=l*l&g[y+=l][x+=l]<'!':x=y=l););z=[...g[y]],z[x]=c,g[y]=z.join``},l=s.length,g=Array(h=l-~l).fill(' '.repeat(h)))&&g.join`
`

Test

f=s=>[...s].map((c,i)=>{for(R=_=>Math.random()*h-l|0;x=R(y=R()),!(i-(l/2|0)?x|y&&x*x+y*y<=l*l&g[y+=l][x+=l]<'!':x=y=l););z=[...g[y]],z[x]=c,g[y]=z.join``},l=s.length,g=Array(h=l-~l).fill(' '.repeat(h)))&&g.join`
`

// Less golfed
U=s=>(
  l=s.length,
  h=l-~l, // l+l+1
  g=Array(h).fill(' '.repeat(h)),
  [...s].map((c,i)=>{
    for(R=_=>Math.random()*h-l|0;
        x=R(y=R()), // set x,y to a random value in range -l ... l
        !(i - (l/2|0) // check if at mid point of input string
          ? x|y && // if not, check x and y must not be both 0
            x*x + y*y <= l*l &  // then check position inside the circle of ray L
            g[y+=l][x+=l] < '!' // then add offset L and check if position is not already used 
          : x=y=l // if at midpoint in input string, x and y have fixed value L
         );
       ); // loop until valid position found
    z = [...g[y]];  // modify string at current position: convert to array ...
    z[x] = c;       // ... set element ...
    g[y] = z.join`` // ... back to string
  }),
  g.join`\n`
)  

setInterval(_=>O.textContent=(f(I.value)),1000)

Word <input id=I value='qwert'><pre id=O></pre>

Ruby, 211 207 203 196 characters

Thanks to edc65 for 4 characters

->(x){x=x.chars
o,b,c=x.size
l=o*2+1
a=Array.new(l){Array.new l,' '}
a[o][o]=x.delete_at o/2
a[b][c]=x.pop if a[b=rand(l)][c=rand(l)]==' '&&(b-o)**2+(c-o)**2<=o*o while x[0]
a.map(&:join).join $/}

Explanation:

->(x){...} define an anonymous function that takes in an argument x

x=x.chars transform x from a string into an array of one-character strings

o,b,c=x.size store the length of the input in o for later use. b and c simply need to be initialized to something, so save 2 characters by attaching to a previous assignment.

l=o*2+1 this is the length/width of the field where all characters could possibly go, also the diameter of the explosion circle.

Array.new(l){Array.new l,' '} make an lxl sized 2D array of space characters.

a[o][o]=x.delete_at o/2 sets the center of the array to the center of the values of x (the input), while deleting that value from x

... while x[0] run the block (in this case, the code before while because it is inline) over and over until x is empty. In ruby, accessing an index that does not exist returns nil, which is a falsey value.

a[b=rand(l)][c=rand(l)]==' ' Assign b and c to random values where 0 <= n < l. Then check if the spot at b,c is empty (aka is set to space character)

(b-o)**2+(c-o)**2<=o*o Pythagorean distance check. o is the length of the input. ** is ruby's exponentiation operator, and val<=o*o is shorter than val**0.5<=o.

a[b][c]=x.pop delete the last value from x. Set the position a,b to that value in array a

a[b][c]=x.pop if a[b=rand(l)][c=rand(l)]==' '&&(b-o)**2+(c-o)**2<=o*o while x[0] Set a random position to the last value if that position is free and is within the explosion radius; keep doing this until we run out of characters to place.

$/ is set to the operating system's newline. It's also shorter than "\n"

a.map(&:join).join $/ Map all the arrays in a to a single-string version of themselves (eg ['a','b','c'] -> 'abc'). Take that new array and join it with newlines. Implicit return.