## Precalculus (6th Edition) Blitzer

The graph of the equation ${{x}^{2}}+{{y}^{2}}=100$ symmetric to the x-axis, y-axis and origin.
Consider the equation, ${{x}^{2}}+{{y}^{2}}=100$ Check symmetry about the y-axis: An equation is symmetric about the y-axis if $-x$ is substituted in the function and the result is an equivalent equation then the graph of the equation is symmetric with respect to the y-axis. Substitute $x=-x$ in the equation, \begin{align} & {{\left( -x \right)}^{2}}+{{y}^{2}}=100 \\ & {{x}^{2}}+{{y}^{2}}=100 \end{align} Therefore, the equation is symmetric about the y-axis Now, check symmetry about the x-axis: An equation is symmetric about the x-axis, if $-y$ is substituted in the function and it leads to an equivalent equation than the graph is symmetric with respect to the x-axis. Substitute $y=-y$ , Therefore, \begin{align} & {{x}^{2}}+{{\left( -y \right)}^{2}}=100 \\ & {{x}^{2}}+{{y}^{2}}=100 \end{align} Therefore, the equation is symmetric about the x-axis. Now, check symmetry about the origin: An equation is symmetric about origin if $x=-x,y=-y$ and it leads to an equivalent equation, this implies that the function is symmetric about the origin. Substitute $x=-x,y=-y$ \begin{align} & {{\left( -x \right)}^{2}}+{{\left( -y \right)}^{2}}=100 \\ & {{x}^{2}}+{{y}^{2}}=100 \end{align} Therefore, the equation is symmetric about the origin Hence, the equation is symmetric about the x-axis, y-axis and origin.