AP EAMCET · Maths · Circle
The circle \(\mathrm{S} \equiv x^2+y^2-2 x-4 y+1=0\) cuts the \(y\)-axis at \(\mathrm{A}, \mathrm{B}(\mathrm{OA}\gt\mathrm{OB})\). If the radical axis of \(\mathrm{S} \equiv \mathrm{O}\) and \(\mathrm{S}^{\prime} \equiv x^2\) \(+y^2-4 x-2 y+4=0\) cuts the \(y\)-axis at \(C\) then the ratio in which C divides AB is
- A \(7+2 \sqrt{3}:-7+2 \sqrt{3}\)
- B \(\sqrt{3}+2: \sqrt{3}-2\)
- C \(6-2 \sqrt{3}: 2 \sqrt{3}-6\)
- D \(-3: \sqrt{3}\)
Answer & Solution
Correct Answer
(A) \(7+2 \sqrt{3}:-7+2 \sqrt{3}\)
Step-by-step Solution
Detailed explanation
Given, \(S \equiv x^2+y^2-2 x-4 y+1=0\) If \(x=0 \Rightarrow y^2-4 y+1=0\) \(\Rightarrow y=\frac{4 \pm \sqrt{16-4}}{2}=2 \pm \sqrt{3}\) So, \(A(0,2+\sqrt{3})\) and \(B(0,2-\sqrt{3})\) Now, equation of radical axis is \(S-S^{\prime}=0\) \(\Rightarrow 2 x-2 y-3=0\) If…
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