What is (x+3/2)^2+(y-1)^2= 17/4 ?

The solution to (x+3/2)^2+(y-1)^2= 17/4 is Circle with (a,b)=(-3/2 ,1),r=sqrt(17/4)

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(x+3/2)^2+(y-1)^2= 17/4

(x + \frac{3}{2})^{2} + (y - 1)^{2} = \frac{17}{4}

(a, b) = (- \frac{3}{2}, 1), r = \frac{17}{4}

Solution steps

(x + \frac{3}{2})^{2} + (y - 1)^{2} = \frac{17}{4}

Rewrite

(x + \frac{3}{2})^{2} + (y - 1)^{2} = \frac{17}{4}

in the form of the standard circle equation

(x - (- \frac{3}{2}))^{2} + (y - 1)^{2} = (\frac{17}{4})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{3}{2}, 1), r = \frac{17}{4}

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What is (x+3/2)^2+(y-1)^2= 17/4 ?
The solution to (x+3/2)^2+(y-1)^2= 17/4 is Circle with (a,b)=(-3/2 ,1),r=sqrt(17/4)