MHT CET · Maths · Circle
If the circles \(x^{2}+y^{2}+2 x+2 k y+6=0\), \(x^{2}+y^{2}+2 k y+k=0\) intersect orthogonally, then \(k\) is
- A 2 or \(-3 / 2\)
- B \(-2\) or \(-3 / 2\)
- C 2 or \(3 / 2\)
- D \(-2\) or \(3 / 2\)
Answer & Solution
Correct Answer
(A) 2 or \(-3 / 2\)
Step-by-step Solution
Detailed explanation
Since, circles cut orthogonally, if
\(2 g g^{\prime}+2 f f^{\prime}=c+c^{\prime} \)
\( \Rightarrow 0+2 k^{2}=6+k \)
\( \Rightarrow 2 k^{2}-k-6=0 \)
\( \Rightarrow (2 k+3)(k-2)=0\)
\(\Rightarrow k=2\) or \(-3 / 2\)
\(2 g g^{\prime}+2 f f^{\prime}=c+c^{\prime} \)
\( \Rightarrow 0+2 k^{2}=6+k \)
\( \Rightarrow 2 k^{2}-k-6=0 \)
\( \Rightarrow (2 k+3)(k-2)=0\)
\(\Rightarrow k=2\) or \(-3 / 2\)
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