# Find the centre and the radius of the circle. [closed]

Find the centre and the radius of the circle.

$$4x^2+4y^2-4x-8y-11=0$$

Thank you.

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Show us what you have tried so we can help you. –  Alizter Aug 19 '13 at 14:32
$$(4x^2+4y^2-4x-8y-11=0) \cdot \left(\frac {1}{4}\right)$$ $$x^2+y^2-x-2y-{\frac{11}{4}}=0$$ $${\text{compared to: }}x^2+y^2+2gx+2fy-c=0$$ $$2g=-1 → g= -{\frac{1}{2}}$$ $$2f=-2 → f= -1$$ $$(-g,-f)=({\frac{1}{2}}, 1)$$ $${\text{This is where I get my answer wrong: }} r^2 = g^2+f^2-c → r = -\sqrt{\frac{3}{2}} \\$$ $${\text{and thank you, Sami! }}$$ –  Audrey G Aug 19 '13 at 15:05
You can simply complete the squares after multiplying by $\frac14$: you get $$\left(x-\frac12\right)^2+(y-1)^2-\frac{11}4=\frac14+1\;,$$ so $$\left(x-\frac12\right)^2+(y-1)^2=\frac32\;.$$ This immediately puts the centre at $\left\langle\frac12,1\right\rangle$ and the radius at $\sqrt{\frac32}$. Note, though, that the radius must be positive: that rules out $-\sqrt{\frac32}$. –  Brian M. Scott Aug 19 '13 at 18:29

## closed as off-topic by Daniel Rust, Hagen von Eitzen, Amzoti, user1337, O.L.Aug 19 '13 at 15:00

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By completing the square $$4x^2+4y^2-4x-8y-11=0\iff (2x-1)^2+(2y-2)^2=4^2\\\iff\left(x-\frac{1}{2}\right)^2+(y-1)^2=4$$