Official Solution:How many points of intersection does the curve \(x^2 + y^2 = 4\) have with line \(x + y = 2\) ?

A. 0

B. 1

C. 2

D. 3

E. 4

Curve \(x^2 + y^2 = 4\) is a circle with radius 2 and the center at the origin. A line cannot have more than 2 points of intersection with a circle, so we can eliminate choices D and E. To answer the question, we can either draw a sketch or solve the system of equations. The second approach gives \((2 - y)^2 + y^2 = 4\) or \(2y^2 - 4y + 4 = 4\) or \(y^2 - 2y = 0\) from where \(y = 0\) and \(y = 2\). Thus, the line and the circle intersect at points \((2, 0)\) and \((0, 2)\).

Alternative Explanation Look at the diagram below:

\(x^2 + y^2 = 4\) is the equation of a circle centered at the origin and with the radius of 2. Now, \(y=2-x\) is the equation of a line with \(x\)-intercept \((2, 0)\) and \(y\)-intercept \((0,2)\). Notice that these points are also the intercepts of given circle with \(X\) and \(Y\) axis hence at these points the line and the circle intersect each other.

Answer: C

I did not understand this part "\((2 - y)^2 + y^2 = 4\) or \(2y^2 - 4y + 4 = 4\) or \(y^2 - 2y = 0\) from where \(y = 0\) and \(y = 2\). Thus, the line and the circle intersect at points \((2, 0)\) and \((0, 2)\)." like where did this come from "[m](2 - y)^2 + y^2 = 4" ?