There is probably some simple geometric way of doing this - it can't be coincidence that the ray passes through the centre of the circle and that the terminus of the ray lies on this circle.

But a blunt apporach is:

Get the Cartesian equation of , easy if you recognise it as a circle of radius 5 and centre at (8, 6).

Get the equation of the line that the ray lies on. Note that the gradient of this line is and it passes through the point (4, 3).

Get the intersection of the circle and the line. The solution you want to keep is the one such that y > 3 (a diagram makes this clear).