What is 40(y-5)^2+40(x+2)^2=2000 ?

The solution to 40(y-5)^2+40(x+2)^2=2000 is Circle with (a,b)=(-2,5),r=5sqrt(2)

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40(y-5)^2+40(x+2)^2=2000

40 (y - 5)^{2} + 40 (x + 2)^{2} = 2000

(a, b) = (- 2, 5), r = 52

Solution steps

40 (y - 5)^{2} + 40 (x + 2)^{2} = 2000

Rewrite

40 (y - 5)^{2} + 40 (x + 2)^{2} = 2000

in the form of the standard circle equation

(x - (- 2))^{2} + (y - 5)^{2} = (52)^{2}

Therefore the circle properties are:

(a, b) = (- 2, 5), r = 52

5 x^{2} + 5 y^{2} - 3 x = 0

x^{2} + y^{2} + 4 y - 21 = 0

(x - 4)^{2} + (y - 0)^{2} = 16

y^{2} = (7 + x) (7 - x)

x^{2} + y^{2} - 4 x - 4 y - 8 = 0

What is 40(y-5)^2+40(x+2)^2=2000 ?
The solution to 40(y-5)^2+40(x+2)^2=2000 is Circle with (a,b)=(-2,5),r=5sqrt(2)