What is (x+6)^2+(y-10)^2=397 ?

The solution to (x+6)^2+(y-10)^2=397 is Circle with (a,b)=(-6,10),r=sqrt(397)

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(x+6)^2+(y-10)^2=397

(x + 6)^{2} + (y - 10)^{2} = 397

(a, b) = (- 6, 10), r = 397

Solution steps

(x + 6)^{2} + (y - 10)^{2} = 397

Rewrite

(x + 6)^{2} + (y - 10)^{2} = 397

in the form of the standard circle equation

(x - (- 6))^{2} + (y - 10)^{2} = (397)^{2}

Therefore the circle properties are:

(a, b) = (- 6, 10), r = 397

(x - 2)^{2} + (y + 6)^{2} = 25 x

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

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

(x + 4)^{2} + (y - 1)^{2} - 9 = 17

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

What is (x+6)^2+(y-10)^2=397 ?
The solution to (x+6)^2+(y-10)^2=397 is Circle with (a,b)=(-6,10),r=sqrt(397)