What is (x-3)^2+(y-6)^2=17 ?

The solution to (x-3)^2+(y-6)^2=17 is Circle with (a,b)=(3,6),r=sqrt(17)

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(x-3)^2+(y-6)^2=17

(x - 3)^{2} + (y - 6)^{2} = 17

(a, b) = (3, 6), r = 17

Solution steps

(x - 3)^{2} + (y - 6)^{2} = 17

Rewrite

(x - 3)^{2} + (y - 6)^{2} = 17

in the form of the standard circle equation

(x - 3)^{2} + (y - 6)^{2} = (17)^{2}

Therefore the circle properties are:

(a, b) = (3, 6), r = 17

x^{2} + y^{2} - 2 x - 2 y + 1 = 0

x^{2} + z^{2} = 4

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

a^{2} + b^{2} = 5

x^{2} + y^{2} - 6 y - 16 = 0

What is (x-3)^2+(y-6)^2=17 ?
The solution to (x-3)^2+(y-6)^2=17 is Circle with (a,b)=(3,6),r=sqrt(17)