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

First use the standard form:

`(x-h)^2+(y-k)^2=r^2`
(h,k) - the coordiantes of the center

The center is (-2,1).

`(x-(-2))^2+(y-1)^2=r^2 \\ (x+2)^2+(y-1)^2=r^2`

The circle passes through the point (x,y)=(-4,1).

`(-4+2)^2+(1-1)^2=r^2 \\ (-2)^2+0^2=r^2 \\ r^2=4`

The standard form is:

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

Convert it to the general form:

`(x+2)^2+(y-1)^2=4 \\ x^2+4x+4+y^2-2y+1=4 \\ x^2+y^2+4x-2y+4+1-4=0 \\ \boxed{x^2+y^2+4x-2y+1=0}`