What is (x^2+y^2-4x-9)/4 =0 ?

The solution to (x^2+y^2-4x-9)/4 =0 is Circle with (a,b)=(2,0),r=sqrt(13)

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(x^2+y^2-4x-9)/4 =0

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

(a, b) = (2, 0), r = 13

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - 2)^{2} + (y - 0)^{2} = (13)^{2}

Therefore the circle properties are:

(a, b) = (2, 0), r = 13

x^{2} + y^{2} - 1.09 x - 7.64 y = 58.18

2 x - x^{2} - y^{2} \geq 0

16 x^{2} + 16 y^{2} + 8 x - 32 y = 1

\frac{( x - 3 ) ^{2}}{25} + \frac{( y + 1 ) ^{2}}{25} = 1

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

What is (x^2+y^2-4x-9)/4 =0 ?
The solution to (x^2+y^2-4x-9)/4 =0 is Circle with (a,b)=(2,0),r=sqrt(13)