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

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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.
The relation does not define $$ y $$ as a function of $$ x $$ because for some $$ x $$-values, there are two corresponding $$ y $$-values.

Determine if the relation defines as a function of

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