When approaching intersection problems, it is important to have both equations and solve them by substituting one into the other or simultaneously. First we need to find the equation of the line l since we already have the equation of the circle c. The equation of a line follows the standard format y=mx+c and we are told that the gradient is 3 (m=3) and the intercept is at (0,1) so c=1. So the equation of the line is y=3x+1. Then, we substitute this equation into the equation of the circle (into the y^2) and expand the (3x+1)^2. We collect similar terms and arrive at 10x^2+6x=0 and then factorise the x to give the solutions x=0 and x=-3/5. Once we have the x values, it is easier to substitute these back into the equation of line l to derive our values for y. So the points of intersection are (0,1) and (-3/5,-4/5).