Question: Complete the equation of the circle $x^2 + y^2 + 2x - 6y + 6 = 0$ in standard form. Identify the center and radius, then select the correct graph of the circle. $(x + 1)^2 + (y - 3)^2 = 4$ The center is $C(-1, 3)$ and the radius is $r = 4$. Graph shapes: A: circle centered near $(1, -3)$, radius about $2$. B: circle centered near $(1, 3)$, radius about $2$. C: circle centered below the x-axis, partially visible. D: circle centered near $(-1, 3)$, radius about $2$. position_hint: centerUser:

1. **State the problem:** We are given the circle equation $$x^2 + y^2 + 2x - 6y + 6 = 0$$ and need to write it in standard form, identify the center and radius, and select the correct graph. 2. **Recall 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. **Complete the square for $x$ and $y$ terms:** Group $x$ and $y$ terms: $$ x^2 + 2x + y^2 - 6y = -6 $$ Complete the square for $x$: $$ x^2 + 2x = (x^2 + 2x + 1) - 1 = (x + 1)^2 - 1 $$ Complete the square for $y$: $$ y^2 - 6y = (y^2 - 6y + 9) - 9 = (y - 3)^2 - 9 $$ 4. **Rewrite the equation with completed squares:** $$ (x + 1)^2 - 1 + (y - 3)^2 - 9 = -6 $$ Simplify the constants: $$ (x + 1)^2 + (y - 3)^2 - 10 = -6 $$ Add $10$ to both sides: $$ (x + 1)^2 + (y - 3)^2 = 4 $$ 5. **Identify the center and radius:** Center: $C(-1, 3)$ Radius: $r = \sqrt{4} = 2$ 6. **Select the correct graph:** The circle is centered at $(-1, 3)$ with radius $2$, so the correct graph is **D**. **Final answer:** Standard form: $$ (x + 1)^2 + (y - 3)^2 = 4 $$ Center: $(-1, 3)$ Radius: $2$ Graph: D