Problem 28 - Precalculus Essentials

Question

Determine whether the graph of each equation is symmetric with respect to the \(y\) -axis, the \(x\) -axis, the origin, more than one of these, or none of these. $$x^{2}+y^{2}=49$$

Step-by-Step Solution

Answer

The graph of the equation \(x^{2}+y^{2}=49\) is symmetric with respect to the y-axis, the x-axis, and the origin.

Step 1: Symmetry with respect to the Y-axis

Replace every \(x\) in the equation with \(-x\) and see if the resulting equation is equivalent to the original. Doing so on our equation yields \((-x)^{2}+y^{2}=49\), simplifying to \(x^{2}+y^{2}=49\). This is the original equation, so it is symmetric with respect to the y-axis.

Step 2: Symmetry with respect to the X-axis

Replace \(y\) with \(-y\) and see if the resulting equation is equivalent to the original. This gives us \(x^{2}+(-y)^{2}=49\), simplifying to \(x^{2}+y^{2}=49\). This is the original equation, so the graph is also symmetric with respect to the x-axis.

Step 3: Symmetry with respect to the Origin

This graph doesn't produce a different shape if you turn it \(180^{\circ}\) around the origin, nor does the graph produce different shapes under reflection along the X and Y axes. Replace both \(x\) and \(y\) with \(-x\) and \(-y\) in the equation respectively. This gives us \((-x)^{2}+(-y)^{2}=49\), simplifying to \(x^{2}+y^{2}=49\). As this is our original equation, it means the graph is symmetric with respect to the origin.