What is x^2+y^2-8x-10y+40=0 ?

The solution to x^2+y^2-8x-10y+40=0 is Circle with (a,b)=(4,5),r=1

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x^2+y^2-8x-10y+40=0

x^{2} + y^{2} - 8 x - 10 y + 40 = 0

(a, b) = (4, 5), r = 1

Solution steps

x^{2} + y^{2} - 8 x - 10 y + 40 = 0

Rewrite

x^{2} + y^{2} - 8 x - 10 y + 40 = 0

in the form of the standard circle equation

(x - 4)^{2} + (y - 5)^{2} = 1^{2}

Therefore the circle properties are:

(a, b) = (4, 5), r = 1

x^{2} + y^{2} - 18 x - 14 y + 66 = 0

x^{2} + y^{2} - 4 x + 8 y + 11 = 0

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

x^{2} - 4 x + y^{2} + 6 y - 24 = 0

(x - 2)^{2} + y^{2} = 36

What is x^2+y^2-8x-10y+40=0 ?
The solution to x^2+y^2-8x-10y+40=0 is Circle with (a,b)=(4,5),r=1