AP EAMCET · Maths · Circle
If \(\mathrm{A}\) and \(\mathrm{B}\) are the points of intersection of the circles \(x^2+y^2-4 x+6 y-3=0\) and \(x^2+y^2+2 x-2 y-2=0\) then the distance between \(\mathrm{A}\) and \(\mathrm{B}\) is
- A \(\frac{13}{10}\)
- B \(\frac{\sqrt{41}}{3}\)
- C \(\frac{\sqrt{231}}{5}\)
- D \(\frac{26}{5}\)
Answer & Solution
Correct Answer
(C) \(\frac{\sqrt{231}}{5}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text {} S_1: x^2+y^2-4 x+6 y-3=0.....(i) \\ & S_2: x^2+y^2+2 x-2 y-2=0.....(ii)\end{aligned}\) \[ \begin{aligned} & S_1:(x-2)^2+(y+3)^2=4^2 \\ & S_2:(x+1)^2+(y-1)^2=2^2 \end{aligned} \] Here,…
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