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What is the general form of the equation of a circle with its center at [tex]\((-2,1)\)[/tex] and passing through [tex]\((-4,1)\)[/tex]?

A. [tex]\(x^2 + y^2 - 4x + 2y + 1 = 0\)[/tex]

B. [tex]\(2^2 + 2 \cdot 4 + 2y + 1 = 0\)[/tex]

C. [tex]\(x^2 + y^2 + 4x - 2y + 9 = 0\)[/tex]

D. [tex]\(x^2 - 2x + y + 4 = 0\)[/tex]


Sagot :

To determine the general form of the equation of a circle with center [tex]\((-2, 1)\)[/tex] that passes through the point [tex]\((-4, 1)\)[/tex], we can follow a series of steps.

### Step-by-Step Solution:

1. Circle’s Standard Form:
The standard form of the equation of a circle is:
[tex]\[ (x - h)^2 + (y - k)^2 = r^2 \][/tex]
where [tex]\((h, k)\)[/tex] is the center of the circle and [tex]\(r\)[/tex] is the radius.

2. Identify the Center:
Given the center [tex]\((-2, 1)\)[/tex], we have:
[tex]\[ h = -2 \quad \text{and} \quad k = 1 \][/tex]

3. Find the Radius:
The radius [tex]\(r\)[/tex] is the distance between the center [tex]\((-2, 1)\)[/tex] and the given point [tex]\((-4, 1)\)[/tex]. The distance formula is:
[tex]\[ r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Substituting the given points:
[tex]\[ r = \sqrt{(-4 - (-2))^2 + (1 - 1)^2} = \sqrt{(-4 + 2)^2} = \sqrt{(-2)^2} = \sqrt{4} = 2 \][/tex]

4. Substitute Center and Radius into Standard Form:
Substituting [tex]\( h = -2 \)[/tex], [tex]\( k = 1 \)[/tex], and [tex]\( r = 2 \)[/tex] into the standard form:
[tex]\[ (x + 2)^2 + (y - 1)^2 = 4 \][/tex]

5. Expand the Equation:
[tex]\[ (x + 2)^2 = x^2 + 4x + 4 \][/tex]
[tex]\[ (y - 1)^2 = y^2 - 2y + 1 \][/tex]
Putting these together:
[tex]\[ x^2 + 4x + 4 + y^2 - 2y + 1 = 4 \][/tex]

6. Simplify:
Combine like terms:
[tex]\[ x^2 + y^2 + 4x - 2y + 5 = 4 \][/tex]
Subtract [tex]\(4\)[/tex] from both sides to get the general form:
[tex]\[ x^2 + y^2 + 4x - 2y + 1 = 0 \][/tex]

### Conclusion:

The correct answer is:
[tex]\[ A. \quad x^2 + y^2 + 4x - 2y + 1 = 0 \][/tex]