What is (x+3.2625400000000…)^2+(y+8)^2=+1936 ?

The solution to (x+3.2625400000000…)^2+(y+8)^2=+1936 is Circle with (a,b)=(-3.26254…,-8),r=44

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(x+3.2625400000000…)^2+(y+8)^2=+1936

(x + 3.2625400000000 \dots)^{2} + (y + 8)^{2} = + 1936

(a, b) = (- 3.26254 \dots, - 8), r = 44

Solution steps

(x + 3.26254 \dots)^{2} + (y + 8)^{2} = 1936

Rewrite

(x + 3.26254 \dots)^{2} + (y + 8)^{2} = + 1936

in the form of the standard circle equation

(x - (- 3.26254 \dots))^{2} + (y - (- 8))^{2} = 4 4^{2}

Therefore the circle properties are:

(a, b) = (- 3.26254 \dots, - 8), r = 44

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

x^{2} + y^{2} - 22 x + 12 y - 150 = 0

(x + 1)^{2} + (y - 5)^{2} = 9

x^{2} + y^{2} + 20 x - 18 y + 145 = 0

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

What is (x+3.2625400000000…)^2+(y+8)^2=+1936 ?
The solution to (x+3.2625400000000…)^2+(y+8)^2=+1936 is Circle with (a,b)=(-3.26254…,-8),r=44