Question

analyze a student draws a circle on a coordinate plane. the center of the circle is at the origin. is this circle a function or a relation? explain.

Explanation:

A function is a relation where each input (x - value) has exactly one output (y - value). For a circle centered at the origin with equation \(x^{2}+y^{2}=r^{2}\) (where \(r\) is the radius), if we solve for \(y\), we get \(y=\pm\sqrt{r^{2}-x^{2}}\). For a given non - zero \(x\) value (where \(|x|\lt r\)), there are two \(y\) values. A relation is a set of ordered pairs. Since a circle is a set of ordered pairs \((x,y)\) that satisfy \(x^{2}+y^{2}=r^{2}\), it is a relation. But it does not meet the criteria of a function.

Answer:

The circle is a relation. A function requires that for each \(x\) - value there is exactly one \(y\) - value. For a circle centered at the origin (equation \(x^{2}+y^{2}=r^{2}\)), for most \(x\) values (where \(|x|\lt r\)), there are two \(y\) values (\(y = \sqrt{r^{2}-x^{2}}\) and \(y=-\sqrt{r^{2}-x^{2}}\)). So it is not a function. Since a relation is just a set of ordered pairs \((x,y)\) (and the circle is a set of ordered pairs that satisfy \(x^{2}+y^{2}=r^{2}\)), it is a relation.