Cho : (C1) : $x^2 + y^2 -2x + 4y +1 = 0 $
         (C2) : $ x^2+y^2+4x-4y+4 = 0$
Lập phương trình tiếp tuyến chung 
Ta có $(C_1): I_1(1;-2) ;R_1=2$
$(C_2): I_2(2;-2); R_2=2$
Nhận xét có $R_1=R_2$
$I_1I_2=1 < R \Rightarrow (C_1), (C_2)$ cắt nhau
$\Rightarrow$ Số tiếp tuyên chung là hai
Mặt khác nhận thấy $(I_1I_2): y=-2 //Ox$
$\Rightarrow$ Tiếp tuyến chung $\triangle : y=c $
ta có $d_{\triangle-I_1}=d_{\triangle-I_2}=2$
$\Rightarrow |-2-c|=2$
$\Rightarrow c=0$
Hoặc $c=-4$
Vậy hai tiếp tuyến chung là $y=0 , y=-4$
Nếu đúng bạn nhấn V để chấp nhận, nhấn mũi tên để vote up cho mình nha. Cảm ơn :) –  NguyễnTốngKhánhLinh 03-07-13 08:37 AM

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