What is (x-(-1))^2+(y-4)^2=2^2 ?

The solution to (x-(-1))^2+(y-4)^2=2^2 is Circle with (a,b)=(-1,4),r=2

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(x-(-1))^2+(y-4)^2=2^2

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

(a, b) = (- 1, 4), r = 2

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 1, 4), r = 2

x, x^{2} - 6 x + Y (x)^{2} + 4 Y = 36

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

1 = 16 - x^{2} - y^{2}

x^{2} + y^{2} - 8 x + 18 y - 3 = 0

x^{2} + y^{2} + 10 x + 8 y + 1 = 0

What is (x-(-1))^2+(y-4)^2=2^2 ?
The solution to (x-(-1))^2+(y-4)^2=2^2 is Circle with (a,b)=(-1,4),r=2