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Rewrite the equation of the circle (x-2)^2+y^2=3 in general form
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@dude @Vocaloid @Narad
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|dw:1558148886282:dw| expand (x-2)^2 w/ foil, arrange all the terms in the equation in the order described above
@dude x^2-4x+y^2-5=0??
@Narad
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It's the reverse of the 1st question, Complete the squares \[x ^{2}+ax+y ^{2}+by+c=0\] \[x^{2}+ax+a^2/4 +y ^{2}+by+b^2/4=-c+a^2/4+b^2/4\] \[(x+a/2)^2+(y+b/2)^2=-c+a^2/4+b^2/4\] If b doesn't exists, then b=0
The -5 is not wright
@Vocaloid I got x^2-4x+y^2+1=0?
Yes
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