Respuesta :

[tex](x-(-2))^2+(y-1)^2=3^2\\(x+2)^2+y^2-2y+1-9=0\\x^2+4x+4+y^2-2y-8=0\\x^2+y^2+4x-2y-4=0[/tex]

Answer:

x²+4x+y²−2y−4=0

Step-by-step explanation:

we can observe that,

1) the coordinates of the center of the circle (h,k) = (-2,1)

2) the radius, r = 3

using the standard form of the equation of circle:

(x−h)²+(y−k)²=r²  (substitute h= -2, k=1 and r=3), we get

(x+2)²+(y−1)²=9  ----> this is in "Standard" form, to get "General" form, simply expand the parenthesis reduce to simplest terms.

(x+2)²+(y−1)²=9

[x² + (2)(x)(2) + 2²] + [y² + (2)(y)(-1) + (-1)²] = 9 (expand and reduce)

x²+4x+y²−2y−4=0 (answer)