Question

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

Answer 1

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 2

(x - h)^2 + (y - k)^2 = r^2 ; (h,k) = (-2,1) and r is the distance between (-2,1) and (-4,1)

So then plug in the values given, expand the squared expressions and subtract r^2 from both sides of the equation to get your equation in general form.

Answer 3

Distance between (-2,1) and (-4,1) is 2. (x - (-2))^2 + (y - 1)^2 = (2)^2 -->

(x+2)^2 + (y-1)^2 = 4 --> x^2 + 4x + 4 + y^2 - 2y + 1 = 4 -->

x^2 + 4x + y^2 - 2y + 5 = 4 --> x^2 + 4x + y^2 - 2y + 1 = 0

Answer 4

Or x^2 + y^2 + 4x - 2y + 1 = 0