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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

X^{2} + Y^{2} + 40 X - 10 Y + 55 = 0

x^{2} + y^{2} + 6 x - 10 y - 17 = 0

(x + 5)^{2} + (y + 7)^{2} = 2 1^{2}

x^{2} + y^{2} - 4225 = 0

y 3. 6^{2} = (x + 0.001)^{2} + (y + 0.3)^{2}

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