Given that the equation of a circle is x ^ 2 + y ^ 2 + 4x-2y + 3 = 0, then the center coordinates and radius of the circle are

Given that the equation of a circle is x ^ 2 + y ^ 2 + 4x-2y + 3 = 0, then the center coordinates and radius of the circle are

formula
x²+4x+4+y²-2y+1=-3+4+1
(x+2)²+(y-1)²=2
So the center of the circle (- 2,1)
r=√2

Let a = {x | x ^ 2-ax-5 = 0}, - 5 ∈ a, then the sum of all elements in the set B = {x | x ^ 2-4x-a = 0} is?
Let a = {x | x ^ 2-ax-5 = 0}, - 5 ∈ a, then the sum of all elements in the set B = {x | x ^ 2-4x-a = 0} is________

-5 ∈ a, substituting x = - 5 into the equation in a: 25 + 5a-5 = 0, we get a = - 4
Then B = {x | x ^ 2-4x + 4 = 0} = {x | (X-2) ^ 2 = 0} = {2}
That is, the sum of all elements in B is 2

Let - 5 ∈ {x | x2-ax-5 = 0}, then the sum of all elements in the set {x | x2-4x-a = 0} is______ ．

Because - 5 ∈ {x | x2-ax-5 = 0}, so 25 + 5a-5 = 0, so a = - 4, x2-4x-a = 0, that is, x2-4x + 4 = 0, the solution is x = 2, so the set {x | x2-4x-a = 0} = {2}. The sum of all elements in the set {x | x2-4x-a = 0} is: 2