Укажите центр и радиус окружности заданной уравнением: (x+4)^2+(y+3)^2=12
Помогите пожалуйста.

радиус=√12

(-4;-3)-центрРадиус=√12

(x+3)2+(y−4)2=12(x+3)2+(y-4)2=12Move all variables to the left side and all constants to the right side.(x+3)2+(y−4)2=12(x+3)2+(y-4)2=12This is the form of a circle. Use this form to determine the center and radius of the circle.(x−h)2+(y−k)2=r2(x-h)2+(y-k)2=r2Match the values in this circle to those of the standard form. The variable rr represents the radius of the circle, hh represents the x-offset from the origin, and kk represents the y-offset from origin.r=2√3r=23h=−3h=-3k=4k=4The center of the circle is found at (h,k)(h,k).Center: (−3,4)(-3,4)These values represent the important values for graphing and analyzing a circle.Center: (−3,4)(-3,4)Radius: 2√3