JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
Let \(C: x^2+y^2=4\) and \(C^{\prime}: x^2+y^2-4 \lambda x+9=0\) be two circles. If the set of all values of \(\lambda\) so that the circles \(\mathrm{C}\) and \(\mathrm{C}^{\prime}\) intersect at two distinct points, is \({R}-[a, b]\), then the point \((8 a+12,16 b-20)\) lies on the curve :
- A \(x^2+2 y^2-5 x+6 y=3\)
- B \(5 x^2-y=-11\)
- C \(x^2-4 y^2=7\)
- D \(6 x^2+y^2=42\)
Answer & Solution
Correct Answer
(D) \(6 x^2+y^2=42\)
Step-by-step Solution
Detailed explanation
\(x^2+y^2=4\) \(C(0,0) \quad \quad r_1=2\) \(C^{\prime}(2 \lambda, 0) \quad r_2=\sqrt{4 \lambda^2-9}\) \(\left|\mathrm{r}_1-\mathrm{r}_2\right|<\mathrm{CC} \mathrm{C}^{\prime}<\left|\mathrm{r}_1+\mathrm{r}_2\right|\)…
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