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What is the center of a circle whose equation is [tex]x^2 + y^2 + 4x - 8y + 11 = 0[/tex]?

A. [tex](-2, 4)[/tex]
B. [tex](-4, 8)[/tex]
C. [tex](2, -4)[/tex]
D. [tex](4, -8)[/tex]


Sagot :

To determine the center of the circle given by the equation [tex]\( x^2 + y^2 + 4x - 8y + 11 = 0 \)[/tex], we will rewrite this equation in the standard form of a circle’s equation, which is [tex]\( (x - h)^2 + (y - k)^2 = r^2 \)[/tex]. This standard form allows us to identify the center [tex]\((h, k)\)[/tex] directly.

### Steps to Rewrite the Equation

1. Group the [tex]\(x\)[/tex] terms and the [tex]\(y\)[/tex] terms:
[tex]\[ x^2 + 4x + y^2 - 8y = -11 \][/tex]

2. Complete the square for the [tex]\(x\)[/tex] terms.
- For [tex]\(x^2 + 4x\)[/tex], take half the coefficient of [tex]\(x\)[/tex] (which is 4), square it, and add inside the square while balancing it outside:
[tex]\[ x^2 + 4x \implies x^2 + 4x + 4 - 4 = (x + 2)^2 - 4 \][/tex]

3. Complete the square for the [tex]\(y\)[/tex] terms.
- For [tex]\(y^2 - 8y\)[/tex], take half the coefficient of [tex]\(y\)[/tex] (which is -8), square it, and add inside the square while balancing it outside:
[tex]\[ y^2 - 8y \implies y^2 - 8y + 16 - 16 = (y - 4)^2 - 16 \][/tex]

4. Combine these completed squares back into the equation.
[tex]\[ (x + 2)^2 - 4 + (y - 4)^2 - 16 = -11 \][/tex]

5. Simplify by combining constants:
[tex]\[ (x + 2)^2 + (y - 4)^2 - 20 = -11 \implies (x + 2)^2 + (y - 4)^2 = 9 \][/tex]

The equation [tex]\((x + 2)^2 + (y - 4)^2 = 9\)[/tex] is now in the standard form of a circle, where [tex]\((h, k)\)[/tex] is the center of the circle, and [tex]\(r^2 = 9\)[/tex], so the radius is 3.

From this standard form, we can identify the center of the circle [tex]\((h, k)\)[/tex]:

### Center of the Circle
[tex]\[ (h, k) = (-2, 4) \][/tex]

Thus, the center of the circle given by the equation [tex]\( x^2 + y^2 + 4x - 8y + 11 = 0 \)[/tex] is [tex]\(\boxed{(-2, 4)}\)[/tex].