What is 10(x+5)^2+10(y-8^2)^2=100 ?

The solution to 10(x+5)^2+10(y-8^2)^2=100 is Circle with (a,b)=(-5,64),r=sqrt(10)

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10(x+5)^2+10(y-8^2)^2=100

10 (x + 5)^{2} + 10 (y - 8^{2})^{2} = 100

(a, b) = (- 5, 64), r = 10

Solution steps

10 (x + 5)^{2} + 10 (y - 8^{2})^{2} = 100

Rewrite

10 (x + 5)^{2} + 10 (y - 8^{2})^{2} = 100

in the form of the standard circle equation

(x - (- 5))^{2} + (y - 64)^{2} = (10)^{2}

Therefore the circle properties are:

(a, b) = (- 5, 64), r = 10

(x + 3)^{2} + (y - 4)^{2} = 75

x^{2} + y^{2} - 4 x + 3 y = 0

x^{2} + y^{2} - 6 x + 4 y - 11 = 0

74 = (24 - x)^{2} + (22 - y)^{2}

(x + 1)^{2} + (y - 1)^{2} = 9 ((x - 2)^{2} + (y + 1)^{2})

What is 10(x+5)^2+10(y-8^2)^2=100 ?
The solution to 10(x+5)^2+10(y-8^2)^2=100 is Circle with (a,b)=(-5,64),r=sqrt(10)