(b) $$ (x-2)^{2}+(y+1)^{2}=9 $$ center $$ (x, y)=(\square) $$ radius $$ \quad r=\square $$

Question

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For the given equation of the circle:

$$
(x - 2)^{2} + (y + 1)^{2} = 9
$$

This is in the standard form of a circle's equation:

$$
(x - h)^{2} + (y - k)^{2} = r^{2}
$$

Where:
- $$(h, k)$$ is the center of the circle.
- $$r$$ is the radius.

**Identifying the center $$(h, k)$$ and the radius $$r$$:**
- $$h = 2$$
- $$k = -1$$
- $$r^{2} = 9 \Rightarrow r = 3$$

**Filled in values:**
- **Center:** $$(x, y) = (2, -1)$$
- **Radius:** $$r = 3$$

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**Answer:**

• Center $$(x, y) = (2,\ -1)$$

• Radius $$ \, r = 3 $$
Center $$(x, y) = (2, -1)$$, radius $$ r = 3 $$.

(b) \( (x-2)^{2}+(y+1)^{2}=9 \) center \( (x, y)=(\square) \) radius \( \quad r=\square \)