Dany jest okrąg o równaniu x^2-2x+y^2+12y+27=0. oblicz r
x²-2x+y²+12y+27=0
(x-1)²+(y+6)²-1-36+27=0
(x-1)²+(y+6)²-10=0
(x-1)²+(y+6)²=10
(x-1)²+(y+6)²=(√10)²
Stąd:
r=√10
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x²-2x+y²+12y+27=0
(x-1)²+(y+6)²-1-36+27=0
(x-1)²+(y+6)²-10=0
(x-1)²+(y+6)²=10
(x-1)²+(y+6)²=(√10)²
Stąd:
r=√10