What is (x-6)^2+(y+3)^2=10 ?

The solution to (x-6)^2+(y+3)^2=10 is Circle with (a,b)=(6,-3),r=sqrt(10)

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(x-6)^2+(y+3)^2=10

(x - 6)^{2} + (y + 3)^{2} = 10

(a, b) = (6, - 3), r = 10

Solution steps

(x - 6)^{2} + (y + 3)^{2} = 10

Rewrite

(x - 6)^{2} + (y + 3)^{2} = 10

in the form of the standard circle equation

(x - 6)^{2} + (y - (- 3))^{2} = (10)^{2}

Therefore the circle properties are:

(a, b) = (6, - 3), r = 10

(x + 4)^{2} + (y - 3)^{2} = 36

x^{2} + y^{2} - 14 x - 12 y + 76 = 0

\frac{( x - 3 ) ^{2}}{4} + \frac{( y + 2 ) ^{2}}{4} = 4

8 x^{2} + 8 y^{2} - 32 x + 64 y - 376 = 0

4 x^{2} + 4 y^{2} - 4 x + 8 y + 1 = 0

What is (x-6)^2+(y+3)^2=10 ?
The solution to (x-6)^2+(y+3)^2=10 is Circle with (a,b)=(6,-3),r=sqrt(10)