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.