What is (x-(-3))^2+(y-(-2))^2=8^2 ?

The solution to (x-(-3))^2+(y-(-2))^2=8^2 is Circle with (a,b)=(-3,-2),r=8

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(x-(-3))^2+(y-(-2))^2=8^2

(x - (- 3))^{2} + (y - (- 2))^{2} = 8^{2}

(a, b) = (- 3, - 2), r = 8

Solution steps

(x - (- 3))^{2} + (y - (- 2))^{2} = 8^{2}

Rewrite

(x - (- 3))^{2} + (y - (- 2))^{2} = 8^{2}

in the form of the standard circle equation

(x - (- 3))^{2} + (y - (- 2))^{2} = 8^{2}

Therefore the circle properties are:

(a, b) = (- 3, - 2), r = 8

(a + bi) \cdot (a - bi) = 1

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

\frac{x ^{2} + y ^{2}}{2} = 1

(x + \frac{1}{2})^{2} + (y - \frac{3}{4})^{2} = 4

(x - 3)^{2} + (y - 2)^{2} = 13

What is (x-(-3))^2+(y-(-2))^2=8^2 ?
The solution to (x-(-3))^2+(y-(-2))^2=8^2 is Circle with (a,b)=(-3,-2),r=8