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Gọi $(C)$ có tâm $I(a;b)$ bán kính R$IA^2=(a-1)^2 + (b-2)^2$$IB^2=(a-3)^2 + (b-4)^2$$d(I,d)=I(3a+b-3I)/ \sqrt{1} $$\begin{cases}IA^{2}=IB^{2} \\ IA^{}= d(I,d)\end{cases}$$\begin{cases}a+b=5 \\ 10(a^{2}+b^{2} -2a-4b+5)= \left| {3a+b-3} \right|\end{cases}$$\begin{cases}a=5-b \\ b=\frac{7}{2} hoặc b=1 \end{cases}$$a=\frac{3}{2} ; b=\frac{7}{2} \Rightarrow (C): (x-\frac{3}{2} )^2 + (y-\frac{7}{2} )^2 = \frac{\sqrt{10} }{2} $$a= 4 ; b=1 \Rightarrow (C): (x-4)^2 + (y-1)^2 = \sqrt{10} $

Gọi $(C)$ có tâm $I(a;b)$ bán kính R$IA^2=(a-1)^2 + (b-2)^2$$IB^2=(a-3)^2 + (b-4)^2$$d(I,d)=I(3a+b-3I)/ \sqrt{10} $$\begin{cases}IA^{2}=IB^{2} \\ IA^{}= d(I,d)\end{cases}$$\begin{cases}a+b=5 \\ 10(a^{2}+b^{2} -2a-4b+5)= \left| {3a+b-3} \right|\end{cases}$$\begin{cases}a=5-b \\ b=\frac{7}{2} hoặc b=1 \end{cases}$$a=\frac{3}{2} ; b=\frac{7}{2} \Rightarrow (C): (x-\frac{3}{2} )^2 + (y-\frac{7}{2} )^2 = \frac{\sqrt{10} }{2} $$a= 4 ; b=1 \Rightarrow (C): (x-4)^2 + (y-1)^2 = \sqrt{10} $

Gọi (C) có tâm I(a;b) bán kính RIA^2=(a-1)^2 + (b-2)^2IB^2=(a-3)^2 + (b-4)^2d(I,d)=I3a+b-3I/căn 1\begin{cases}IA^{2}=IB^{2} \\ IA^{}= d(I,d)\end{cases}\begin{cases}a+b=5 \\ 10(a^{2}+b^{2} -2a-4b+5)= \left| {3a+b-3} \right|\end{cases}\begin{cases}a=5-b \\ b=\frac{7}{2} hoặc b=1 \end{cases}a=3/2 b=7/2 => (C): (x-3/2)^2 + (y-7/2)^2 = (căn 10)/2a=4 b=1 => (C): (x-4)^2 + (y-1)^2 = căn 10

Gọi $(C)$ có tâm $I(a;b)$ bán kính R$IA^2=(a-1)^2 + (b-2)^2$$IB^2=(a-3)^2 + (b-4)^2$$d(I,d)=I(3a+b-3I)/ \sqrt{1} $$\begin{cases}IA^{2}=IB^{2} \\ IA^{}= d(I,d)\end{cases}$$\begin{cases}a+b=5 \\ 10(a^{2}+b^{2} -2a-4b+5)= \left| {3a+b-3} \right|\end{cases}$$\begin{cases}a=5-b \\ b=\frac{7}{2} hoặc b=1 \end{cases}$$a=\frac{3}{2} ; b=\frac{7}{2} \Rightarrow (C): (x-\frac{3}{2} )^2 + (y-\frac{7}{2} )^2 = \frac{\sqrt{10} }{2} $$a= 4 ; b=1 \Rightarrow (C): (x-4)^2 + (y-1)^2 = \sqrt{10} $