What is (x+6)^2+(y-10)^2=12 ?

The solution to (x+6)^2+(y-10)^2=12 is Circle with (a,b)=(-6,10),r=sqrt(12)

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(x+6)^2+(y-10)^2=12

(x + 6)^{2} + (y - 10)^{2} = 12

(a, b) = (- 6, 10), r = 12

Solution steps

(x + 6)^{2} + (y - 10)^{2} = 12

Rewrite

(x + 6)^{2} + (y - 10)^{2} = 12

in the form of the standard circle equation

(x - (- 6))^{2} + (y - 10)^{2} = (12)^{2}

Therefore the circle properties are:

(a, b) = (- 6, 10), r = 12

(x - (- 2))^{2} + (y - (- 1))^{2} = (3)^{2}

x^{2} + y^{2} - 9 x + 4 y - 138 = 0

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

x^{2} - 6 x + y^{2} + 8 y + 5 = 0

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

What is (x+6)^2+(y-10)^2=12 ?
The solution to (x+6)^2+(y-10)^2=12 is Circle with (a,b)=(-6,10),r=sqrt(12)