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}

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)