What is 4(x^2-2x+1)+4(y^2-y+0)=11+1+0 ?

The solution to 4(x^2-2x+1)+4(y^2-y+0)=11+1+0 is Circle with (a,b)=(1, 1/2),r=(sqrt(13))/2

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4(x^2-2x+1)+4(y^2-y+0)=11+1+0

4 (x^{2} - 2 x + 1) + 4 (y^{2} - y + 0) = 11 + 1 + 0

(a, b) = (1, \frac{1}{2}), r = \frac{13}{2}

Solution steps

4 (x^{2} - 2 x + 1) + 4 (y^{2} - y + 0) = 11 + 1 + 0

Rewrite

4 (x^{2} - 2 x + 1) + 4 (y^{2} - y + 0) = 11 + 1 + 0

in the form of the standard circle equation

(x - 1)^{2} + (y - \frac{1}{2})^{2} = (\frac{13}{2})^{2}

Therefore the circle properties are:

(a, b) = (1, \frac{1}{2}), r = \frac{13}{2}

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x^{2} + y^{2} - 8 x - 14 y - 7 = 0

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What is 4(x^2-2x+1)+4(y^2-y+0)=11+1+0 ?
The solution to 4(x^2-2x+1)+4(y^2-y+0)=11+1+0 is Circle with (a,b)=(1, 1/2),r=(sqrt(13))/2