What is y^2+x^2+24x+10y+160=0 ?

The solution to y^2+x^2+24x+10y+160=0 is Circle with (a,b)=(-12,-5),r=3

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y^2+x^2+24x+10y+160=0

y^{2} + x^{2} + 24 x + 10 y + 160 = 0

(a, b) = (- 12, - 5), r = 3

Solution steps

y^{2} + x^{2} + 24 x + 10 y + 160 = 0

Rewrite

y^{2} + x^{2} + 24 x + 10 y + 160 = 0

in the form of the standard circle equation

(x - (- 12))^{2} + (y - (- 5))^{2} = 3^{2}

Therefore the circle properties are:

(a, b) = (- 12, - 5), r = 3

16 (x^{2} + y^{2}) = 16 (14 x + 6 y + 6)

(x + 4)^{2} + (y - 2)^{2} = \frac{( 8 ^{2} )}{( 5 )}

x^{2} + y^{2} + 5 = 2 x + 8 y + 10

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

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

What is y^2+x^2+24x+10y+160=0 ?
The solution to y^2+x^2+24x+10y+160=0 is Circle with (a,b)=(-12,-5),r=3