1. The problem is to analyze the given equation of a circle: $$(x + 3)^2 + (y + 5)^2 = 25$$ 2. This equation is in the standard form of a circle: $$(x - h)^2 + (y - k)^2 = r^2$$ where $(h, k)$ is the center and $r$ is the radius. 3. By comparing, we identify the center as:\ $$(-3, -5)$$ 4. The radius squared is 25, so the radius is:\ $$r = \sqrt{25} = 5$$ 5. Therefore, the circle has center $(-3, -5)$ and radius 5. 6. The intercepts can be found by setting $x=0$ or $y=0$: - For $x=0$, substitute into the equation: $$ (0+3)^2 + (y+5)^2 = 25 \Rightarrow 9 + (y+5)^2 = 25 \Rightarrow (y+5)^2 = 16 $$ $$ y + 5 = \pm 4 \Rightarrow y = -5 \pm 4 $$ So, $y = -1$ or $y = -9$ giving points $(0, -1)$ and $(0, -9)$. - For $y=0$, substitute into the equation: $$ (x+3)^2 + (0+5)^2 = 25 \Rightarrow (x+3)^2 + 25 = 25 \Rightarrow (x+3)^2 = 0 $$ $$ x + 3 = 0 \Rightarrow x = -3 $$ So, the only x-intercept is $(-3,0)$. Final answer: The circle has center $(-3, -5)$ and radius $5$. Its x-intercept is $(-3, 0)$, and y-intercepts are $(0,-1)$ and $(0,-9)$.