JEE Mains · Maths · STD 11 - 4.1 complex nubers
Let a circle \(C\) in complex plane pass tltrough the points \(z _{1}=3+4 i , z _{2}=4+3 i\) and \(z _{3}=5 i\). If \(z \left(\neq z _{1}\right)\) is a point on \(C\) such that the line through \(z\) and \(z _{1}\) is perpendicular to the line through \(z _{2}\) and \(z _{3}\), then \(\arg ( z )\) is equal to
- A \(\tan ^{-1}\left(\frac{2}{\sqrt{5}}\right)-\pi\)
- B \(\tan ^{-1}\left(\frac{24}{7}\right)-\pi\)
- C \(\tan ^{-1}(3)-\pi\)
- D \(\tan ^{-1}\left(\frac{3}{4}\right)-\pi\)
Answer & Solution
Correct Answer
(B) \(\tan ^{-1}\left(\frac{24}{7}\right)-\pi\)
Step-by-step Solution
Detailed explanation
Slope of \(BC =\frac{3-5}{4-0}=-\frac{1}{2}\) Slope of \(AP =2\) equation of AP : \(y-4=2(x-3)\) \(\Rightarrow y =2( x -1)\) \(P\) lies on circle \(x^{2}+y^{2}=25\) \(\Rightarrow x ^{2}+(2( x -1))^{2}=25\) \(\Rightarrow x =-\frac{7}{5}\) and \(y =-\frac{24}{5}\)…
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