JEE Mains · Maths · STD 11 - 4.1 complex nubers
The number of points of intersection of \(| z -(4+3 i )|=2\) and \(| z |+| z -4|=6, z \in C\) is
- A \(0\)
- B \(1\)
- C \(2\)
- D \(3\)
Answer & Solution
Correct Answer
(C) \(2\)
Step-by-step Solution
Detailed explanation
\(C:(x-4)^{2}+(y-3)^{2}=4\) \(E: \frac{(x-2)^{2}}{9}+\frac{y^{2}}{5}=1\) Lower Extremity of vertical diameter of circle \(\rightarrow(4,1)\) Put in ellipse \(\Rightarrow \frac{(4-2)^{2}}{9}+\frac{1}{5}-1\) \(=\frac{4}{9}+\frac{1}{5}-1\) \(=\frac{29}{45}-1<0\) Two Solutions…
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