y2 +8y + 16 = 0 can factor to (y+4) (y+4) = 0 so y+4 = 0 so y = -4
x2+y2+4x+2y+3=0(x+2)2 + (Y+2.5)2 = 3This is the equation of circle with center at (-2,-2.5) with radius 3.5
Given: x2 + y2 - 10x + 4y + 4 = 0 First, we'll move our constants to the right: x2 + y2 - 10x + 4y = -4 Then group terms with the same variables together: x2 - 10x + y2 + 4y = -4 Then complete the squares: x2 - 10x + 25 + y2 + 4y + 4 = -4 + 25 + 4 (x - 5)2 + (y + 2)2 = 25 And there we have it. This is an equation for a circle whose center point is at (5, -2), with a radius of √25, which equals 5.
A=0 b=0 c=0
11 = sqrt of 121. it is a circle centred on the origin think what would happen on the line x=0 (The y axis) the equation simplifies to y2 = 121 or y =11 you can also think of eqn of a circle as x2+y2=r2
yes that is correct
No.
Yes
(0, -14)
(2-r)e-rr
y2 +8y + 16 = 0 can factor to (y+4) (y+4) = 0 so y+4 = 0 so y = -4
dont know sorry
Center is (0, 0) . . . the origin.Radius is 7.
If y2 + 13y = 40, then y ~ 15.55If y2 + 13y = 0, then y = -13Improved Answer:-y2+13y+40 = 0(y+5)(y+8) = 0Therefore: y = -5 and y = -8
Yes, as x-y2=0
The center of the circle is at (0, 0) and its radius is the square root of 1 which is 1
x2 + y2 - 5y + 4 = 0 x2 + y2 - 5y = - 4 x2 + y2 - 5y + 25/4 = - 4 + 25/4 = 9/4 x2 + (y - 5/2)2 = (3/2)2 Centre = (0, 5/2) and radius = 3/2