AP EAMCET · Maths · Circle
If the cosine of the angle between the two circles \(x^2+y^2+2 x+4 y-3=0\) and \(x^2+y^2+2 k x-2 y-1=0\) is \(\frac{1}{2 \sqrt{3}}\) then \(k^2=\)
- A 2
- B 4
- C 16
- D 8
Answer & Solution
Correct Answer
(B) 4
Step-by-step Solution
Detailed explanation
\(r_1^2 = 1^2+2^2-(-3) = 8\) \(r_2^2 = k^2+(-1)^2-(-1) = k^2+2\) \(d^2 = (-k-(-1))^2+(1-(-2))^2 = (1-k)^2+3^2 = k^2-2k+10\) \(\cos \theta = \frac{|r_1^2+r_2^2-d^2|}{2r_1r_2} \implies \frac{1}{2\sqrt{3}} = \frac{|8+(k^2+2)-(k^2-2k+10)|}{2\sqrt{8}\sqrt{k^2+2}}\)…
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