i) Using implicit differentiation find dy/dx for x^2 + y^2 = 4 ii) At what points is the tangent to the curve parallel to the y axis iii) Given the line y=x+c only intersects the circle once find c given that c is positive.

I don't know how effectively I can communicate this answer via text without the whiteboard but I'll try.

i) First implicitly differentiate with respect to x: 2x + 2y * dy/dx = 0.

Rearranging gives dy/dx = -x/y. I would estimate this to be worth 3 marks

ii) The curve is parallel to the y axis when dx/dy = 0 or equivalently when dy/dx tends to inf. This can be seen by the denominator in our fraction for dy/dx tending to 0 => y=0. For y=0 we have two valid points from substituting back into the equation of the circle, solving the quadratic in x gives the two points (2,0) and (-2,0). (3 marks)

ii) The line will only intersect the circle once iff at the intersection point the value of dy/dx matches for both the line and the circle (A diagram would help clear this up). dy/dx for the line is 1 and substituting that into our result found in i) it is found that y= -x. Substitute that result back into the equation of the cirlce and you find that x^2 = 2. This pair of equations has 2 solutions (sqrt(2), -sqrt(2)), (-sqrt(2), sqrt(2)). From a sketch you would be able to see that the second of these points must be chosen for a positive c and subsituting that point into the equation of a straight line you find that c = 2*sqrt(2). (5 marks)