Question 6647
Your problem is in step 2, where you solved for y by taking the square root of both sides of the equation.  You must include a "+ or -" symbol in this step, where the plus or the minus is determined by the point that is selected.  What you have is the equation of a circle, and the point of tangency is at (-3,4) placing the point in the second quadrant, since x is negative and y is positive.  Notice that in the second quadrant, the slope of a tangent line to a point on the curve will have a positive slope (by inspection!), so you have to use the plus sign for the slope of the tangent line.  In quadrant IV, where x is positive and y is negative, you also have a positive slope.  In quadrants I and III, the tangent line to a circle will  have a negative slope (again by inspection!).


I think the rest of what you have done is correct.


R^2 at SCC