What is (X+4)^2+(Y+1)^2=25 ?

The solution to (X+4)^2+(Y+1)^2=25 is Circle with (a,b)=(-4,-1),r=5

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(X+4)^2+(Y+1)^2=25

(X + 4)^{2} + (Y + 1)^{2} = 25

(a, b) = (- 4, - 1), r = 5

Solution steps

(X + 4)^{2} + (Y + 1)^{2} = 25

Rewrite

(X + 4)^{2} + (Y + 1)^{2} = 25

in the form of the standard circle equation

(X - (- 4))^{2} + (Y - (- 1))^{2} = 5^{2}

Therefore the circle properties are:

(a, b) = (- 4, - 1), r = 5

(x - 3)^{2} + (y + 4)^{2} = 5

(x - 3)^{2} + (y + 4)^{2} = 3

4 x^{2} + 12 x + 4 y^{2} + 16 - 11 = 0

x^{2} = 50 - y^{2}

215 = a^{2} + b^{2}

What is (X+4)^2+(Y+1)^2=25 ?
The solution to (X+4)^2+(Y+1)^2=25 is Circle with (a,b)=(-4,-1),r=5