What is (x-19)^2+(y+18)^2=289 ?

The solution to (x-19)^2+(y+18)^2=289 is Circle with (a,b)=(19,-18),r=17

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(x-19)^2+(y+18)^2=289

(x - 19)^{2} + (y + 18)^{2} = 289

(a, b) = (19, - 18), r = 17

Solution steps

(x - 19)^{2} + (y + 18)^{2} = 289

Rewrite

(x - 19)^{2} + (y + 18)^{2} = 289

in the form of the standard circle equation

(x - 19)^{2} + (y - (- 18))^{2} = 1 7^{2}

Therefore the circle properties are:

(a, b) = (19, - 18), r = 17

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

2 x^{2} - 20 x + 2 y^{2} + 12 y = - 18

(x - 5)^{2} + y^{2} - 8 y = 0

x^{2} + y^{2} - 8 x + 4 y = 0

(x + 4)^{2} + (y - 4 \frac{1}{2})^{2} = 42 \frac{1}{4}

What is (x-19)^2+(y+18)^2=289 ?
The solution to (x-19)^2+(y+18)^2=289 is Circle with (a,b)=(19,-18),r=17