Determine the point of intersection between a circle (x+2)^2 + (y-2)^2 =1 and the parabola y=-(x+2)^2 +3.can anyone please outline the steps to solve it I have tried solve by substitution but I get Stuck

Answer 1

just plug and chug.
(x+2)^2 + (y-2)^2 =1
so (x+2)^2 = 1-(y-2)^2

y = -(x+2)^2+3
y = (y-2)^1-1 + 3
y = y^2-4y+4+2
y^2-5y+6 = 0
(y-3)(y-2) = 0
So the curves intersect at (-2,3) and (-2±1,2)

http://www.wolframalpha.com/input/?i=plot+(x%2B2)%5E2+%2B+(y-2)%5E2+%3D1,+y%3D-(x%2B2)%5E2+%2B3

Answer 2

Thanks Steve you saved my day