Select the correct answer.
What is the general form of the equation of a circle with its center at (-2, 1) and passing through (-4, 1)?

A.
x2 + y2 − 4x + 2y + 1 = 0

B.
x2 + y2 + 4x − 2y + 1 = 0

C.
x2 + y2 + 4x − 2y + 9 = 0

D.
x2 − y2 + 2x + y + 1 = 0


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Answer :

Answer:

[tex]x^{2} + y^{2} + 4\, x - 2\, y + 1 = 0[/tex].

Step-by-step explanation:

Approach this problem in the following steps:

  • Find the radius of the circle from the coordinates of the center of this circle and that of a point on this circle.
  • Obtain the standard form equation of this circle.
  • Simplify and rewrite the standard form equation to obtain the general form equation in the requested format.

The distance between the center of a circle and any point on the circle is equal to the radius of the circle. Given the coordinates of the center [tex](x_{0},\, y_{0})[/tex] and a point [tex](x_{1},\, y_{1})[/tex] on this circle, radius of the circle would be equal to the distance between these two points:

[tex]\begin{aligned}r &= \sqrt{(x_{1} - x_{0})^{2} + (y_{1} - y_{0})^{2}} \\&= \sqrt{((-4) - (-2))^{2} - (1 - 1)^{2}} \\ &= 2\end{aligned}[/tex].

The standard form equation of a circle of radius [tex]r[/tex] and center [tex](x_{0},\, y_{0})[/tex] is:

[tex](x - x_{0})^{2} + (y - y_{0})^{2} = r^{2}[/tex].

Since [tex]r = 2[/tex] for the circle in this question, the standard form equation of this circle would be:

[tex](x - (-2))^{2} + (y - (1))^{2} = (2)^{2}[/tex].

Rearrange the standard form equation to obtain an equivalent equation for this circle in the required format:

[tex](x + 2)^{2} + (y - 1)^{2} = 2^{2}[/tex].

[tex]x^{2} + y^{2} + 4\, x - 2\, y + 1 = 0[/tex].

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