The first stage in solving this problem, of course, is to draw a diagram of the situation under consideration, such as the one below:

In this problem, we assume that the rabbit starts its run up the y-axis from the origin, at time t=0.  The dog starts its run from the point (c,0) at time t=0.  We further assume that the dog always runs “towards” the rabbit – that is, at any time t the tangent of the dog’s path passes through the location of the rabbit at time t.  This is shown in the diagram above. 

(it is worth noting that this is why we always draw a diagram – it is usually easier to express these ideas diagrammatically, rather than in words!) 

Now, if the dog is at point (x,y) at some time t, then we can see that the equation of the path of the dog can be expressed as 

In coming up with this expression, we have noted that 

To make progress with this equation, we differentiate with respect to x to give, after slight simplification, 

Now, using the chain rule of differentiation and the geometry in the blue insert in the diagram above, we find that 

and hence 

To solve this equation we set 

and with our initial conditions, this yields 

Solving this and then solving for y (subject to x = c when y = 0) eventually yields 

for the path of the dog.  An example, with k = 0.7 i.e. the rabbit runs at 70% of the speed of the dog), is shown in the diagram below.  The dog starts his run from the point (0,1). 

As to whether the dog can catch the rabbit, well you can see that this all depends on whether y remains finite as x approaches zero.  From the equation for y above, we can see that if k is greater than one, then y tends to infinity as x approaches zero.  Thus, provided the rabbit can run faster than the dog, the dog cannot catch him!