Find the points of intersection of x^2 + y^2 = 4 and x^2 + y^2 - 4x - 4y = -4?

Question

Please give a step-by-step solution!!!

Answer 1

circle x^2 + y^2 = 2^2 has centre (0, 0) and radius 2.

circle x^2 + y^2 -- 4x -- 4y + 4 = 0 has centre (2, 2) and radius 2. intersecting at two points (2, 0) and (0, 2) only given by 4 --4x --4y + 4 = 0

Or x + y = 2 and x^2 + (2 -- x)^2 = 4 whence x = 2, 0 and y = 0, 2.

Answer 2

You are finding the intersection of a circle and an ellipse.  If you plug the equations into each other, you get 4 - 4x -4y = -4.  This simplified gives you the equation y = 2-x.  The solutions to the problem lie on this line.  Plug this equation back into the first equation and solve to find the values.

x^2 + (2-x)^2 = 4

x^2 + 4 - 4x + x^2 = 4

2x^2 - 4x = 0

2x( x - 2) = 0

x = 0,2

Answer 3

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