What is (x-11)^2+(y-6)^2-7=0 ?

The solution to (x-11)^2+(y-6)^2-7=0 is Circle with (a,b)=(11,6),r=sqrt(7)

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(x-11)^2+(y-6)^2-7=0

(x - 11)^{2} + (y - 6)^{2} - 7 = 0

(a, b) = (11, 6), r = 7

Solution steps

(x - 11)^{2} + (y - 6)^{2} - 7 = 0

Rewrite

(x - 11)^{2} + (y - 6)^{2} - 7 = 0

in the form of the standard circle equation

(x - 11)^{2} + (y - 6)^{2} = (7)^{2}

Therefore the circle properties are:

(a, b) = (11, 6), r = 7

(x + 1)^{2} + (y - 3)^{2} = 16

x^{2} + y^{2} < 36

x^{2} + y^{2} < 2 x

x^{2} + y^{2} - 6 x - 8 y + 24 = 0

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

What is (x-11)^2+(y-6)^2-7=0 ?
The solution to (x-11)^2+(y-6)^2-7=0 is Circle with (a,b)=(11,6),r=sqrt(7)