JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
Let \(S_{1}: x^{2}+y^{2}=9\) and \(S_{2}:(x-2)^{2}+y^{2}=1\). Then the locus of center of a variable circle \(S\) which touches \(S_{1}\) internally and \(S_{2}\) externally always passes through the points :
- A \((0, \pm\, \sqrt{3})\)
- B \(\left(\frac{1}{2}, \pm\, \frac{\sqrt{5}}{2}\right)\)
- C \(\left(2,\, \pm\, \frac{3}{2}\right)\)
- D \((1,\pm \,2)\)
Answer & Solution
Correct Answer
(C) \(\left(2,\, \pm\, \frac{3}{2}\right)\)
Step-by-step Solution
Detailed explanation
\(S_{1}: x^{2}+y^{2}=9\) \(9\rightarrow\) \(r_{1}=3\) \(9\rightarrow\) \(A(0,0)\) \(S_{2}:(x-2)^{2}+y^{2}=1\) \(1\rightarrow\) \(r_{2}=1\) \(1\rightarrow\) \(B(2,0)\) \(\because c _{1} c _{2}= r _{1}- r _{2}\) \(\therefore\) given circle are touching internally Let a veriable…
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