Determine if the relation defines \( y \) as a function of \( x \) \[ (x-3)^{2}+(y-5)^{2}=1 \]

The given equation is: \[ (x - 3)^2 + (y - 5)^2 = 1 \] This represents a **circle** with center at \((3, 5)\) and radius \(1\). ### Vertical Line Test To determine if \(y\) is a function of \(x\), we can use the **Vertical Line Test**. This test states that if any vertical line intersects the graph of the relation more than once, then \(y\) is **not** a function of \(x\). For the given circle: 1. **Draw Vertical Lines:** Consider vertical lines passing through different values of \(x\). 2. **Intersection Points:** For most values of \(x\) within the domain of the circle (\(2 \leq x \leq 4\)), a vertical line will intersect the circle at **two** points. For example, let’s solve for \(y\) in terms of \(x\): \[ (y - 5)^2 = 1 - (x - 3)^2 \] \[ y - 5 = \pm \sqrt{1 - (x - 3)^2} \] \[ y = 5 \pm \sqrt{1 - (x - 3)^2} \] This shows that for each \(x\) in the domain \(2 \leq x \leq 4\), there are **two** corresponding values of \(y\). ### Conclusion Since there are multiple \(y\) values for a single \(x\) value within the domain, the relation **does not** define \(y\) as a function of \(x\). **Answer:** No. For some x-values there are two y-values, so y is not uniquely determined by x.