Answer :
To find the center of the circle given by the equation \( x^2 + y^2 - 4x + 2y - 11 = 0 \), we need to rewrite the equation in the standard form of a circle, which is \((x - h)^2 + (y - k)^2 = r^2\). This form allows us to identify the center \((h, k)\) and the radius \(r\) of the circle.
1. Rewrite the equation by completing the square for the \(x\) and \(y\) terms:
Given equation:
[tex]\[ x^2 + y^2 - 4x + 2y - 11 = 0 \][/tex]
Step 1: Group the \(x\) terms and the \(y\) terms:
[tex]\[ (x^2 - 4x) + (y^2 + 2y) = 11 \][/tex]
Step 2: Complete the square for each group:
For \(x\)-terms \(x^2 - 4x\):
[tex]\[ x^2 - 4x \][/tex]
Add and subtract \(4\) (which is \( (4/2)^2 \)):
[tex]\[ x^2 - 4x + 4 - 4 \][/tex]
Rewriting this:
[tex]\[ (x - 2)^2 - 4 \][/tex]
For \(y\)-terms \(y^2 + 2y\):
[tex]\[ y^2 + 2y \][/tex]
Add and subtract \(1\) (which is \((2/2)^2 \)):
[tex]\[ y^2 + 2y + 1 - 1 \][/tex]
Rewriting this:
[tex]\[ (y + 1)^2 - 1 \][/tex]
Step 3: Substitute these completed squares back into the equation:
[tex]\[ (x - 2)^2 - 4 + (y + 1)^2 - 1 = 11 \][/tex]
Step 4: Combine constant terms:
[tex]\[ (x - 2)^2 + (y + 1)^2 - 5 = 11 \][/tex]
[tex]\[ (x - 2)^2 + (y + 1)^2 = 16 \][/tex]
Now, the equation is in the standard form \((x - h)^2 + (y - k)^2 = r^2\), where \( (h, k) \) is the center of the circle and \( r \) is the radius.
2. Identify the center (h, k):
By comparing \((x - 2)^2 + (y + 1)^2 = 16\) with the standard form \((x - h)^2 + (y - k)^2 = r^2\):
- \( h = 2 \)
- \( k = -1 \)
Therefore, the center of the circle is \((2, -1)\).
The correct answer is:
A. [tex]\((2, -1)\)[/tex].
1. Rewrite the equation by completing the square for the \(x\) and \(y\) terms:
Given equation:
[tex]\[ x^2 + y^2 - 4x + 2y - 11 = 0 \][/tex]
Step 1: Group the \(x\) terms and the \(y\) terms:
[tex]\[ (x^2 - 4x) + (y^2 + 2y) = 11 \][/tex]
Step 2: Complete the square for each group:
For \(x\)-terms \(x^2 - 4x\):
[tex]\[ x^2 - 4x \][/tex]
Add and subtract \(4\) (which is \( (4/2)^2 \)):
[tex]\[ x^2 - 4x + 4 - 4 \][/tex]
Rewriting this:
[tex]\[ (x - 2)^2 - 4 \][/tex]
For \(y\)-terms \(y^2 + 2y\):
[tex]\[ y^2 + 2y \][/tex]
Add and subtract \(1\) (which is \((2/2)^2 \)):
[tex]\[ y^2 + 2y + 1 - 1 \][/tex]
Rewriting this:
[tex]\[ (y + 1)^2 - 1 \][/tex]
Step 3: Substitute these completed squares back into the equation:
[tex]\[ (x - 2)^2 - 4 + (y + 1)^2 - 1 = 11 \][/tex]
Step 4: Combine constant terms:
[tex]\[ (x - 2)^2 + (y + 1)^2 - 5 = 11 \][/tex]
[tex]\[ (x - 2)^2 + (y + 1)^2 = 16 \][/tex]
Now, the equation is in the standard form \((x - h)^2 + (y - k)^2 = r^2\), where \( (h, k) \) is the center of the circle and \( r \) is the radius.
2. Identify the center (h, k):
By comparing \((x - 2)^2 + (y + 1)^2 = 16\) with the standard form \((x - h)^2 + (y - k)^2 = r^2\):
- \( h = 2 \)
- \( k = -1 \)
Therefore, the center of the circle is \((2, -1)\).
The correct answer is:
A. [tex]\((2, -1)\)[/tex].
