Question

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

x2 + y2 − 4x + 2y + 1 = 0

x2 + y2 + 4x − 2y + 1 = 0

x2 + y2 + 4x − 2y + 9 = 0

x2 − y2 + 2x + y + 1 = 0

Answer 1

standard form of circle: (x - h)^2 + (y - k)^2 = r^2
center is (h , k) and radius is r

for your circle, the center (h , k) is (-2 , 1)
so it will look like: (x +2)^2 + (y - 1)^2 = r^2

to find the radius squared, plug in the (x,y) point you know is on the circle -- (-4,1)
(-4 +2)^2 + (1 - 1)^2 = r^2
-2^2 + 0 = 4 = r^2

thus, the equation is: (x +2)^2 + (y - 1)^2 = 4

Answer 2

x^2+4+4x+y^2+1-2y=4
x^2+y^2+4x-2y+1=0
so second opton is the answer

Answer 3

are you understand @king12233