TS EAMCET · Maths · Circle
If the number of common tangents to the pair of circles \(x^2+y^2-2 x+4 y-4=0\) and \(x^2+y^2+4 x-4 y+\alpha=0\) is 4 , then the least integral value of \(\alpha\) is
- A 4
- B 5
- C 6
- D 7
Answer & Solution
Correct Answer
(B) 5
Step-by-step Solution
Detailed explanation
We have, \[ \begin{gathered} x^2+y^2-2 x+4 y-4=0 \\ \text { and } x^2+y^2+4 x-4 y+\alpha=0 \\ c_1=(1,-2), r_1=\sqrt{1+4+4}=3 \\ c_2=(-2,2), r_2=\sqrt{4+4-\alpha}=\sqrt{8-\alpha} \end{gathered} \] Circle have 4 common tangents…
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