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

The solution to x^{(2)}+y^{(2)}-2x-4y-5=0 is Circle with (a,b)=(1,2),r=sqrt(10)

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

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

(a, b) = (1, 2), r = 10

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - 1)^{2} + (y - 2)^{2} = (10)^{2}

Therefore the circle properties are:

(a, b) = (1, 2), r = 10

x^{2} + x + y^{2} - 65 y = 1

(x + 4)^{2} + (y - 7)^{2} = 64

x^{2} + y^{2} + 10 x - 2 y + 25 = 0

x^{2} + y^{2} + 4 y = - 3

3 x^{2} + 3 y^{2} + 4 y = 7

What is x^{(2)}+y^{(2)}-2x-4y-5=0 ?
The solution to x^{(2)}+y^{(2)}-2x-4y-5=0 is Circle with (a,b)=(1,2),r=sqrt(10)