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

you need to find the radius of the circle 0 oh r^2 i mean
(
r^2 = (y2 - y1)^2 + (x2-x1)^2 where (x1,y1) and (x2,y2) are the corodinates of the center and the point on the circl

Answer 2

here it is
(-2-(-4))^2 + (1-1)^2

Answer 3

can you work that out/

Answer 4

so they answer I get would be the radius? @welshfella

Answer 5

no it will be the square of the radius which is what we want
- its really pythagoras theorum that we are using

Answer 6

okay. i got 4 for the answer. what would I do to get the answer. like what steps? i'd like to try to understand it.

Answer 7

would the new equation use the square of the radius? like 4^2=(1-1)^2+(-4-2)^2 @welshfella

Answer 8

one year ago
(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 9

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 10

Or you can do x^2 + y^2 + 4x - 2y + 1 = 0

Answer 11

make sense?

Answer 12

so I'd subtract the actual; distance from what part of the equation?

Answer 13

Math is not my best I apologize for not being of any help