A simple collision problem with a not so simple (to me) solution. I could use some help. I know this is a Physics problem and the Physics is fairly simple, but the equation I need to solve seems a bit more difficult than I anticipated. I'm curious if there is a different approach to solving this problem mathematically and/or a more simple solution.
Suppose I have two point particles, Particle 1 is at the origin $(0,0)$ and Particle 2 is at $(x_0, y_0)$. Particle 1 travels at the speed of light $c$ (for this problem we are ignoring special relativity). Particle 2 travels with a constant velocity $(v_{x_0}, v_{y_0})$. We know the particles will collide at some unknown location $(x, y)$. So our goal is to solve for $x$ and $y$. Below is a picture of the situation.

We can use simple kinematic equations to describe the motion of Particle 2: 
$$x = x_0 + v_{x_0}t$$
$$y = y_0 + v_{y_0}t$$
We know that they will collide, which means they must collide at the time it takes Particle 1 to travel to the point $(x,y)$ which is described by: 
$$t = \frac{\sqrt{x^2+y^2}}{c}$$
Plugging this into our previous two equations give us the following:
$$x = x_0 + v_{x_0}\frac{\sqrt{x^2+y^2}}{c}$$
$$y = y_0 + v_{y_0}\frac{\sqrt{x^2+y^2}}{c}$$
I plugged the two equations into Wolfram and I ran out of computing time. I am wondering if there is any other way to go about solving this problem. Is my thought process and final equations correct? If anyone can shed some light on my problem I would be very thankful. 
 A: Instead of trying to compute the collision point directly, compute the collision time (as Rahul commented). Using the distance formula that you used, solve $$(x_0+v_{x_0}t)^2+(y_0+v_{y_0}t)^2 = c^2t^2$$ for $t$, take the least nonnegative solution, and substitute that into the equations of motion of particle 2.
A: Since particle 1 is traveling in an unknown direction, your approach is correct. 
Otherwise it would just be a linear equation.
I would eliminate the square root of your final equation, as it might confuse Wolfram. It might assume complex variables and then simple rules known from real square root cannot be applied anymore.
You get
$\left(\frac{x - x_0}{v_{x_0}/c}\right)^2  = x^2 + y^2$
$\left(\frac{y - y_0}{v_{y_0}/c}\right)^2  = x^2 + y^2$
which is a system of two quadratic equations in two variables. You solve it by solving one equation for the one variable and plugging the result into another. Perhaps Wolfram will give you the solution when you plug in the two equations above, or you do it by hand.
However, you will get multiple solutions, which you need to check if they are correct. This is because the squaring operation introduces new solutions.
A: Take a loot ak the following picture:

Particle 1 starts in $A=(0.0)$, particle 2 starts in $B=(x_0,y_0)$. Particale 2 travels with velocity $\lvert v \rvert$ in the direction given by $v$. Both particles meet at point $C=(x,y)$, which is unknown.
Note that the the value of $\beta=\angle ABC$ is fixed and can be calculated, it is the angle between the lines trough $A$ and $B$ and the velocity vector $v$. If the time it takes for the particles to meet is $t$, we have
$$\overline{AC}=ct, \quad \overline{BC}=\lvert v\rvert t$$
From this follows by the law of sine
$$\frac{\sin \alpha}{\sin{\beta}} = \frac{\overline{BC}}{\overline{AC}} = \frac{\lvert v\rvert}{c},$$
That means $\sin \alpha$ can be calculated by what is given. Under the assumption that $\lvert v\rvert < c$ (a bit of physics), it can be shown that $\alpha$ must be the smaller (< $90°$) of the 2 values from $[0,180°]$ that has that sine value.
Since we now know 2 interior angle values of the triangle $ABC$, we can calculate the remaining one $\gamma$ at $C$ via $\alpha + \beta + \gamma = 180°$, anf finally apply the law of sine again to calculate $\overline{BC}$:
$$\overline{BC}=\overline{AB}\frac{\sin \alpha}{\sin \gamma}$$ and thus
$$ C = B + \overline{BC}\frac{v}{\lvert v \rvert}.$$
