What is (x+2)^2+(y+1/2)^2= 17/4 ?

The solution to (x+2)^2+(y+1/2)^2= 17/4 is Circle with (a,b)=(-2,-1/2),r=sqrt(17/4)

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(x+2)^2+(y+1/2)^2= 17/4

(x + 2)^{2} + (y + \frac{1}{2})^{2} = \frac{17}{4}

(a, b) = (- 2, - \frac{1}{2}), r = \frac{17}{4}

Solution steps

(x + 2)^{2} + (y + \frac{1}{2})^{2} = \frac{17}{4}

Rewrite

(x + 2)^{2} + (y + \frac{1}{2})^{2} = \frac{17}{4}

in the form of the standard circle equation

(x - (- 2))^{2} + (y - (- \frac{1}{2}))^{2} = (\frac{17}{4})^{2}

Therefore the circle properties are:

(a, b) = (- 2, - \frac{1}{2}), r = \frac{17}{4}

(2 x^{2} - 12 x) + (2 y^{2} + 8 y) = 24

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

x^{(2)} + y^{(2)} = \frac{( 9 )}{( 4 )}

x^{2} + 12 x + y^{2} - 16 y = 0

(x - 3)^{2} + (y - 8)^{2} = 36

What is (x+2)^2+(y+1/2)^2= 17/4 ?
The solution to (x+2)^2+(y+1/2)^2= 17/4 is Circle with (a,b)=(-2,-1/2),r=sqrt(17/4)