What is (x-3)^2+(y-4)^2-9=y-x+1 ?

The solution to (x-3)^2+(y-4)^2-9=y-x+1 is Circle with (a,b)=(5/2 , 9/2),r=sqrt(23/2)

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(x-3)^2+(y-4)^2-9=y-x+1

(x - 3)^{2} + (y - 4)^{2} - 9 = y - x + 1

(a, b) = (\frac{5}{2}, \frac{9}{2}), r = \frac{23}{2}

Solution steps

(x - 3)^{2} + (y - 4)^{2} - 9 = y - x + 1

Rewrite

(x - 3)^{2} + (y - 4)^{2} - 9 = y - x + 1

in the form of the standard circle equation

(x - \frac{5}{2})^{2} + (y - \frac{9}{2})^{2} = (\frac{23}{2})^{2}

Therefore the circle properties are:

(a, b) = (\frac{5}{2}, \frac{9}{2}), r = \frac{23}{2}

(x - 1)^{2} + (y + 11)^{2} = 100

9 x^{(2)} + 9 y^{(2)} - 6 x + 24 y = 19

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

(x - 0)^{2} + (y - \frac{1}{2})^{2} = (\frac{1}{2})^{2}

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

What is (x-3)^2+(y-4)^2-9=y-x+1 ?
The solution to (x-3)^2+(y-4)^2-9=y-x+1 is Circle with (a,b)=(5/2 , 9/2),r=sqrt(23/2)