1. **Problem statement:** We analyze the relations given by the equations $$x^{2} + y^{2} = 25$$ and $$y = \\sqrt{25 - x^{2}}$$ to graph them, determine whether they are functions, and state their domain and range. 2. **Analyzing the first relation:** The equation $$x^{2} + y^{2} = 25$$ represents a circle centered at the origin with radius 5. This can be rewritten as $$y^{2} = 25 - x^{2}$$, so for each $$x$$ in the domain, $$y$$ can be $$+\\sqrt{25 - x^{2}}$$ or $$-\\sqrt{25 - x^{2}}$$. 3. **Is the first relation a function?** No, because for some $$x$$ values (e.g., $$x=0$$), there are two corresponding $$y$$ values (positive and negative), violating the vertical line test. 4. **Domain and range of the first relation:** - Domain is all $$x$$ such that $$25 - x^{2} \geq 0$$, thus $$-5 \leq x \leq 5$$. - Range is all $$y$$ such that $$-5 \leq y \leq 5$$ since the circle extends 5 units above and below the x-axis. 5. **Analyzing the second relation:** The equation $$y = \\sqrt{25 - x^{2}}$$ gives only the non-negative root, representing the upper semicircle. 6. **Is the second relation a function?** Yes, for each $$x$$ in the domain, there is exactly one $$y$$. 7. **Domain and range of the second relation:** - Domain is the same as the first: $$-5 \leq x \leq 5$$. - Range is $$0 \leq y \leq 5$$ because $$y$$ is the square root and must be non-negative. **Final answer:** - Relation $$x^{2} + y^{2} = 25$$ is not a function; domain $$[-5,5]$$, range $$[-5,5]$$. - Relation $$y = \\sqrt{25 - x^{2}}$$ is a function; domain $$[-5,5]$$, range $$[0,5]$$.