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$ x^2 + y^2 = 4x+4$

How to put it in the standard form: $(x-a)^2 + (y-b)^2 = r^2$

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1 Answer 1

up vote 5 down vote accepted

$(x-a)^2+(y-b)^2=r^2\implies x^2+y^2-2ax-2by+a^2+b^2-r^2=0$

Comparing with, $x^2+y^2-4x-4=0$

$b=0$ (coefficient of $y$)

$2a=4\implies a=2$ (coefficient of $x$)

$a^2+b^2-r^2=-4,2^2+0^2-r^2=-4,r^2=8$ (constant term)

So, $(x-2)^2+(y-0)^2=(2\sqrt 2)^2$

Or directly, $x^2+y^2-4x-4=(x)^2-2\cdot x\cdot 2+(2)^2+(y)^2-4-4$ $=(x-2)^2+(y)^2-(2\sqrt 2)^2$

So, $(x-2)^2+(y)^2-(2\sqrt 2)^2=0$ or $(x-2)^2+(y)^2=(2\sqrt 2)^2$

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Thank you, I understand. – JohnPhteven Oct 23 '12 at 9:57
@ZafarS, welcome! – lab bhattacharjee Oct 23 '12 at 10:01

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