AP EAMCET · Maths · Circle
The angle between circles \(x^2+y^2+2 x+4 y+1=0\) and \(x^2+y^2-2 x+6 y-3=0\) is
- A \(\cos ^{-1}\left(\frac{3}{\sqrt{13}}\right)\)
- B \(\cos ^{-1}\left(\frac{3}{\sqrt{31}}\right)\)
- C \(\cos ^{-1}\left(\sqrt{\frac{3}{31}}\right)\)
- D \(2 \cos ^{-1}\left(\frac{3}{\sqrt{13}}\right)\)
Answer & Solution
Correct Answer
(A) \(\cos ^{-1}\left(\frac{3}{\sqrt{13}}\right)\)
Step-by-step Solution
Detailed explanation
Circles are \[ \begin{aligned} & c_1, x^2+y^2+2 x+4 y+1=0 \\ & c_2, x^2+y^2-2 x+6 y-3=0 \end{aligned} \] On comparing with general form of circles \(x^2+y^2+2 g x+2 f y+c=0\) where, centre = (− g, − f) and radius…
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