## Precalculus (10th Edition)

(0,2) and ($\sqrt 2,\sqrt 2$) satisfy the equation $x^{2}+y^{2}=4$.
a) If (0,2) lies on the graph of $x^{2}+y^{2}=4$, then it will satisfy the equation. Taking x=0 and y=2, we get- $0^{2}+2^{2}=4$ 4$=$4 Since LHS$=$RHS therefore (0,2) lies on the graph. b) If (-2,2) lies on the graph of $x^{2}+y^{2}=4$, then it will satisfy the equation. Taking x=-2 and y=2, we get- $(-2)^{2}+2^{2}=4$ 8$\ne$4 Since LHS$\ne$RHS therefore (-2,2) doe not lie on the graph. c) If ($\sqrt 2,\sqrt 2$) lies on the graph of $x^{2}+y^{2}=4$, then it will satisfy the equation. Taking x=$\sqrt 2$ and y=$\sqrt 2$, we get- $(\sqrt 2)^{2}+(\sqrt 2)^{2}=4$ 4$=$4 Since LHS$=$RHS therefore ($\sqrt 2,\sqrt 2$) lies on the graph.