What is (x-5)^2+(y+7)^2=4 ?

The solution to (x-5)^2+(y+7)^2=4 is Circle with (a,b)=(5,-7),r=2

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(x-5)^2+(y+7)^2=4

(x - 5)^{2} + (y + 7)^{2} = 4

(a, b) = (5, - 7), r = 2

Solution steps

(x - 5)^{2} + (y + 7)^{2} = 4

Rewrite

(x - 5)^{2} + (y + 7)^{2} = 4

in the form of the standard circle equation

(x - 5)^{2} + (y - (- 7))^{2} = 2^{2}

Therefore the circle properties are:

(a, b) = (5, - 7), r = 2

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

x^{2} + y^{2} = 2 3^{2}

- 2 = x^{2} + y^{2} - 6 x + 5

x^{2} + y^{2} - 4 x - 6 x + 9 = 0

x^{2} + y^{2} + 2 x + 4 y - 12 = 0

What is (x-5)^2+(y+7)^2=4 ?
The solution to (x-5)^2+(y+7)^2=4 is Circle with (a,b)=(5,-7),r=2