Answer:
[tex]x^{2} + y^{2} + 4\, x - 2\, y + 1 = 0[/tex].
Step-by-step explanation:
Approach this problem in the following steps:
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].