TS EAMCET · Maths · Circle
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 \(10 \sqrt{6}\)
- D \(2 \sqrt{6}\)
Answer & Solution
Correct Answer
(D) \(2 \sqrt{6}\)
Step-by-step Solution
Detailed explanation
\(\left(T_1 T\right)^2=\left(C_1 C_2\right)^2-\left(r_2-n_1\right)^2\) \(\begin{aligned} & T T_1=\sqrt{\left(C_1 C_2\right)^2-\left(r_2-\eta_1\right)^2} \\ & C_1 C_2^2=(1+2)^2+(2+2)^2=25\end{aligned}\)…
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