What is x^2+y^2-14x+12y-36=0 ?

The solution to x^2+y^2-14x+12y-36=0 is Circle with (a,b)=(7,-6),r=11

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x^2+y^2-14x+12y-36=0

x^{2} + y^{2} - 14 x + 12 y - 36 = 0

(a, b) = (7, - 6), r = 11

Solution steps

x^{2} + y^{2} - 14 x + 12 y - 36 = 0

Rewrite

x^{2} + y^{2} - 14 x + 12 y - 36 = 0

in the form of the standard circle equation

(x - 7)^{2} + (y - (- 6))^{2} = 1 1^{2}

Therefore the circle properties are:

(a, b) = (7, - 6), r = 11

x^{2} + y^{2} + 6 x - 10 y + 17 = 0

(x - 6)^{2} + (y - 8)^{2} = 1 0^{2}

x^{2} + y^{2} + \frac{4 ( x ^{2} + y ^{2} )}{21} = 25

x^{2} + z^{2} - 3 x - 6 z + 8 = 0

(x - 9)^{2} + (y - 12)^{2} = 4 (x^{2} + y^{2})

What is x^2+y^2-14x+12y-36=0 ?
The solution to x^2+y^2-14x+12y-36=0 is Circle with (a,b)=(7,-6),r=11