What is y^{(2)}+x^{(2)}-4x-5=0 ?

The solution to y^{(2)}+x^{(2)}-4x-5=0 is Circle with (a,b)=(2,0),r=3

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y^{(2)}+x^{(2)}-4x-5=0

y^{(2)} + x^{(2)} - 4 x - 5 = 0

(a, b) = (2, 0), r = 3

Solution steps

y^{2} + x^{2} - 4 x - 5 = 0

Rewrite

y^{2} + x^{2} - 4 x - 5 = 0

in the form of the standard circle equation

(x - 2)^{2} + (y - 0)^{2} = 3^{2}

Therefore the circle properties are:

(a, b) = (2, 0), r = 3

x^{2} + y^{2} = 3 x - y

3 x^{2} + 3 y^{2} - 12 x - 24 y - 48 = 0

x^{2} + y^{2} - 8 x + 4 y - 10 = 0

x^{(2)} + y^{(2)} + 8 x - 6 y = 6

yy = 9 x - x^{2}

What is y^{(2)}+x^{(2)}-4x-5=0 ?
The solution to y^{(2)}+x^{(2)}-4x-5=0 is Circle with (a,b)=(2,0),r=3