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r =2
I.
S1 =(2,-3)
(x -2)^2 + ( y+3)^2 = 4
II.
S2 = (-2,-3)
(x+2)^2 + (y +3)^2 = 4
w 1 przypadku okrag ma rownanie
(x-2)²+(y+3)²=4
w 2 przypadku ma postac
(x+2)²+(y+3)²=4
r =2
S =(2,-3)
(x -2)^2 + ( y+3)^2 = 4
S² = (-2,-3)
(x+2)² + (y +3)² = 4