Answer:
[tex]x^{2}+y^{2} +4x-2y+1=0[/tex]
Step-by-step explanation:
we know that
The general form of the equation of a circle is
[tex]x^{2} +y^{2}+ Dx + Ey + F=0[/tex]
where D, E, F are constants
step 1
Find the radius of the circle
Remember that the distance from the center to any point on the circle is equal to the radius
so
the formula to calculate the distance between two points is equal to
[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]
[tex](-2,1)\\(-4,1)[/tex]
substitute the values
[tex]r=\sqrt{(1-1)^{2}+(-4+2)^{2}}[/tex]
[tex]r=\sqrt{(0)^{2}+(-2)^{2}}[/tex]
[tex]r=2\ units[/tex]
step 2
Find the equation of the circle in standard form
[tex](x-h)^{2} +(y-k)^{2}=r^{2}[/tex]
In this problem we have
center ( -2,1)
radius r=2 units
substitute
[tex](x+2)^{2} +(y-1)^{2}=2^{2}[/tex]
[tex](x+2)^{2} +(y-1)^{2}=4[/tex]
Step 3
Convert to general form
[tex](x+2)^{2} +(y-1)^{2}=4\\ \\x^{2}+4x+4+y^{2}-2y+1=4\\ \\x^{2}+y^{2} +4x-2y+1=0[/tex]
