What is x^2+y^2-8x-20y+52=0 ?

The solution to x^2+y^2-8x-20y+52=0 is Circle with (a,b)=(4,10),r=8

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x^2+y^2-8x-20y+52=0

x^{2} + y^{2} - 8 x - 20 y + 52 = 0

(a, b) = (4, 10), r = 8

Solution steps

x^{2} + y^{2} - 8 x - 20 y + 52 = 0

Rewrite

x^{2} + y^{2} - 8 x - 20 y + 52 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (4, 10), r = 8

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

(x - 2)^{2} + (y - 1)^{2} = 49

y^{2} + x^{2} \leq 1

x^{2} + y^{2} - 6 x - 8 y - 14 = 0

x^{2} + y^{2} + 10 x + 12 y - 3 = 0

What is x^2+y^2-8x-20y+52=0 ?
The solution to x^2+y^2-8x-20y+52=0 is Circle with (a,b)=(4,10),r=8