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

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

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

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

(a, b) = (3, - 2), r = 17

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (3, - 2), r = 17

x^{2} + y^{2} - 12 x = - 4 y + 5

(3 x - 2 - 2)^{(2)} + (3 y + 1)^{(2)} = 9

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

x^{2} + y^{2} - 16 x + 8 y + 64 = 0

4 x^{2} + 4 y^{2} - 6 x + 10 y = \frac{1}{2}

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