WBJEE · Maths · Circle
If four distinct points \((2 k, 3 k),(2,0),(0,3)\) (0,0) lie on a circle, then
- A \(k < 0\)
- B \(0 < k < 1\)
- C \(k=1\)
- D \(k>1\)
Answer & Solution
Correct Answer
(C) \(k=1\)
Step-by-step Solution
Detailed explanation
Since, join of (2,0) and (0,3) subtends \(90^{\circ}\) at (0,0) \(\Rightarrow\) It is a diameter. \(\therefore\) Equation is \((x-2)(x-0)+(y-0)(y-3)=0\) \(x^{2}+y^{2}-2 x-3 y=0\) \((2 k, 3 k)\) lies on it \(\Rightarrow \quad 4 k^{2}+9 k^{2}-4 k-9 k=0\) \(\Rightarrow\) .…
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