What is 25=(x-4)^2+(y+1)^2 ?

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

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25=(x-4)^2+(y+1)^2

25 = (x - 4)^{2} + (y + 1)^{2}

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

Solution steps

25 = (x - 4)^{2} + (y + 1)^{2}

Rewrite

25 = (x - 4)^{2} + (y + 1)^{2}

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 + 1)^{2} + (y - 1)^{2} = 36

x^{2} + y^{2} + 6 x + 10 y + 9 = 0

x^{2} + y^{2} - \frac{2}{3} x + \frac{8}{3} y = \frac{19}{9}

x^{2} + 14 x + y^{2} - 2 y - 50 = 0

(x - 2)^{2} + (y - 1.5)^{2} = 9

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