How do I find the Euclidean distance between one point and all the points in a list?

I would use DistanceMatrix for this. Here is a comparison of DistanceMatrix with the other answers:

SeedRandom[1];
data = RandomReal[1, {10^7, 2}];

r1 = First @ DistanceMatrix[{{1., 1.}}, data]; //AbsoluteTiming
r2 = distance[{1,1}, data]; //AbsoluteTiming (* Mahdi *)
r3 = Outer[EuclideanDistance, {{1., 1.}}, data, 1] //Flatten; //AbsoluteTiming (* RunnyKine *)

r1 == r2 == r3

{0.384095, Null}

{0.88146, Null}

{9.64241, Null}

True

Here is what you want to do. Let the first point's coordinate be stored as a list of list as follows:

x1 = {{6.05102, 5.87667}}    

and the second set of coordinates

y1 = {{1.40687, 4.92494}, {0.419206, 1.70406}, {6.29657,0.577941}, {4.12022, 4.94952},
      {2.04784, 5.94545}, {1.29192,1.43152}, {3.26737, 1.90134}, {4.27274, 0.528028},
      {2.79659,1.37788}, {5.43955, 1.81355}} 

Now to compute the Euclidean distances between x1 and every element in y1 use Outer, your best friend from now on.

Outer[EuclideanDistance, x1, y1, 1]//Flatten

This then gives you

{4.74067, 7.00914, 5.30442, 2.14187, 4.00377, 6.51217, 4.85304, 5.63651, 5.55252, 4.10887}. 

Hope this helps. In fact, you can loop this through various x1's as follows.

Table[Outer[EuclideanDistance, {killer[[k]]}, y1, 1], {k, 1, n}]

Where n is the Length of the killer list. This is fast and compact code.

I understand that this question already has an accepted answer, but couldn't resist posting my answer. This answer is mainly useful if you have a very large dataset. I define the following function using Compile function in Mathematica.

distance = Compile[{{n, _Real, 1}, {z, _Real, 2}},
Sqrt[Total[(# - n)^2]] & /@ z, RuntimeOptions -> "Speed", 
Parallelization -> True, CompilationTarget -> "C", 
RuntimeAttributes -> {Listable}
];

Then we map the function over a two dimensional set with 10000000 (=$10^7$) points.

data = RandomReal[{0, 1}, {10000000, 2}];

The result is:

new[{0, 0}, data] // AbsoluteTiming // First
(* 0.609684 s *)

which is 10 times faster compared to the accepted answer:

Outer[EuclideanDistance, {{0, 0}}, data, 1] // AbsoluteTiming // First
(* 5.927118 s *)