This question was reportedly attempted by majority of the candidates and they performed well in it. Candidates were reported to have first obtained the y – coordinates of the points where x = 0 by substituting 0 for in the given equation to obtain y2 – 6y + 5 = 0. Solving this quadratic equation gave y = 1 and 5. Differentiating 2 + y2 – 2 – 6y + 5 = 0 gave 2+ 2y - 2 - 6 = 0. Bringing like terms together and simplifying gave = . Gradient to the curve at the point (0,1) was at (0,1) = - = The equation of the tangent to the curve x2 + y2 – 2x – 6y + 5 = 0 at (0,1) was y – 1 = (x – 0) i.e. y – 1 = () which gave + 2y – 2 = 0. Similarly, the gradient of the curve at point (0,5) = = = . Therefore, the equation of the tangent at (0,5) = y- 5 = ½ () which gave – 2y + 10 = 0.