If $\mathrm{T}_1 \mathrm{~T}_1{ }^{\prime}$ and $\mathrm{T}_2 \mathrm{~T}_2{ }^{\prime}$ are the common tangents of the circles $\mathrm{S}=\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}-4 \mathrm{y}-4=0$ and $\mathrm{S}^{\prime}=\mathrm{x}^2+\mathrm{y}^2+4 \mathrm{x}+4 \mathrm{y}+$ $4=0$ where $T_1, T_1^{\prime}, T_2, T_2^{\prime}$ are the points of contact, then the distance between $\mathrm{T}_1$ and $\mathrm{T}_1{ }^{\prime}$ is

• A

  $6 \sqrt{6}$
• B

  $5 \sqrt{6}$
• C

  $10 \sqrt{6}$
• Correct answer
  D

  $2 \sqrt{6}$

Solution

$\left(T_1 T\right)^2=\left(C_1 C_2\right)^2-\left(r_2-n_1\right)^2$
$\begin{aligned} & T T_1=\sqrt{\left(C_1 C_2\right)^2-\left(r_2-\eta_1\right)^2} \\ & C_1 C_2^2=(1+2)^2+(2+2)^2=25\end{aligned}$ $\begin{aligned} & \eta=\sqrt{1^2+(-2)^2+4}=3 \\ & n_2=\sqrt{(-2)^2+(-2)^2-4}=2 \\ & \therefore \quad T T_1=\sqrt{25-(1)^2}=\sqrt{24} \\ & \therefore \quad T T_1=2 \sqrt{6} .\end{aligned}$

Hence, the correct answer is option: 'D'.