Đáp án: ` (C') : (x+10/3)^2 + (y-5)^2 = 5/3`
Giải thích các bước giải:
`(C) : I_1 (2;-3) ; R = \sqrt15`
`V_(I;-1/3) : (C) → (C') <=> ` $\begin{cases}I_{1}→I_{2}\\R→R_{2}\\\end{cases}$
Gọi `I_2 (a;b)`
`<=>` $\begin{cases}\overrightarrow{II{2}}=\dfrac{-1}{3} \overrightarrow{II{1}} (1)\\\dfrac{R_{2}}{R_{1}}=|k|=\dfrac{1}{3} (2)\\\end{cases}$
`(1) <=>` $\begin{cases}\overrightarrow{II_{2}} (a+2; b-3)\\ \dfrac{-1}{3} \overrightarrow{II_{1}} (\dfrac{-4}{3} ; 2) \\\end{cases}$
`<=>` $\begin{cases}a+2=\dfrac{-4}{3}\\b-3=2\\\end{cases}$ `<=>` $\begin{cases}a=\dfrac{-10}{3}\\b=5\\\end{cases}$
`=> I_2 (-10/3 ; 5)`
`(2) <=> R_2 = 1/3 R_1 = \sqrt15/3`
`=> (C') : (x+10/3)^2 + (y-5)^2 = 5/3`
