Question
(b) \( (x-2)^{2}+(y+1)^{2}=9 \) center \( (x, y)=(\square) \) radius \( \quad r=\square \)
Ask by Daniel Boone. in the United States
Jan 23,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
Center \((x, y) = (2, -1)\), radius \( r = 3 \).
Solution
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 \( \, r = 3 \)
Reviewed and approved by the UpStudy tutoring team
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Simplify this solution Beyond the Answer
The equation \( (x-2)^{2}+(y+1)^{2}=9 \) represents a circle in standard form. The center of the circle is derived from the values inside the parentheses, which translates to the coordinates \( (2, -1) \). So the center \( (x, y) = (2, -1) \). The radius \( r \) can be found by taking the square root of the right side of the equation. Since \( 9 \) is the square of \( 3 \), we find that \( r = 3 \).
