## Precalculus (6th Edition) Blitzer

Step I: To check symmetry about the y-axis: Put $x=-x$ in the given equation; if the equation remains the same, then it has symmetry about the y-axis. \begin{align} & {{x}^{2}}+{{y}^{2}}=17 \\ & {{\left( -x \right)}^{2}}+{{y}^{2}}=17 \\ & {{x}^{2}}+{{y}^{2}}=17 \end{align} It is the same as the provided equation, hence it has symmetry about the y-axis. Step II: To check symmetry about the x-axis: Put $y=-y$ in the given equation; if the equation remains the same, then it has symmetry about the x-axis. \begin{align} & {{x}^{2}}+{{y}^{2}}=17 \\ & {{x}^{2}}+{{\left( -y \right)}^{2}}=17 \\ & {{x}^{2}}+{{y}^{2}}=17 \end{align} It is the same as the provided equation, hence it has symmetry about the x-axis. Step III: To check symmetry about the origin: Put $x=-x\text{ and }y=-y$ in the given equation; if the equation remains the same, then it has symmetry about the origin. \begin{align} & {{x}^{2}}+{{y}^{2}}=17 \\ & {{\left( -x \right)}^{2}}+{{\left( -y \right)}^{2}}=17 \\ & {{x}^{2}}+{{y}^{2}}=17 \end{align} It is the same as the provided equation, hence it has symmetry about the origin. Therefore, the equation ${{x}^{2}}+{{y}^{2}}=17$ has symmetry about the x-axis, y-axis, and the origin also.