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

using the equation,
(x-h)^2 +(y-k)^2 = r^2

where the center =(h, k)
and the point =(x,y)
inserting,
(x-(-2))^2 + (y- 1)^2 = r^2
(x+2)^2 + (y- 1)^2 = r^2.........................equation
(-4 +2)^2 + (1 -1)^2 = r^2
(-2)^2 + (0)^2 =r^2
4 =r^2

(x+2)^2 + (y- 1)^2 = r^2
(x+2)^2 + (y- 1)^2 = 4
expanding
x2 + 4x +4 +y2 -2y +1 =4
x2 +y2+4x-2y +1 =0

Answer 2

Thank you