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

Step-by-step explanation:

To find the equation of a circle center at (-2,1) and passing through (-4,1), First we need to find its radius

To find its radius, we use (-2, 1)   and (-4, 1)

r²  =  

r²  =  ( -4 - -2)²   +   ( 1-1)²

r²  =  (-4 +2)²  +  0

r²   = (-2)²  

r²  =   4

Equation of a circle is

(x - a)²   +   (y-b)²   = r²

where (a, b) are the center of the circles

From the question;

The circle passes through the center (-2, 1), so our a = -2 and b = 1

We can now substitute our variables into the equation;

(x - a)²   +   (y-b)²   = r²

(x - -2)²  +  (y -1)² =  4

(x+2)²  +   (y-1)²   =4

we can now go ahead and expand the brackets

x² + 4x + 4 + y² -  2y  + 1  = 4

We can rearrange this equation and hence;

x²  + y² + 4x -2y +4   + 1  = 4

x²  + y² + 4x -2y + 5 = 4

x²  + y² + 4x -2y + 5-4 = 0

x²  + y² + 4x -2y  + 1  = 0

12/11c

or

12

c

11

no not all of them.

step-by-step explanation:

some of them have a square base.