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JEE Mains · Maths · STD 11 - 4.1 complex nubers

Let \(z_1\) and \(z_2\) be two complex numbers satisfying \(\left| {{z_1}} \right| = 9\) and \(\left| {{z_2-3-4i}} \right| = 4\).  Then the minimum value of \(\left| {{z_1} - {z_2}} \right|\) is

  1. A \(0\)
  2. B \(\sqrt 2\)
  3. C \(1\)
  4. D \(2\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(0\)

Step-by-step Solution

Detailed explanation

\(|{z_1}|\, = \,9,\) \(|{z_2}\, - \,(3 + 4i)|\, = \,4\) \({C_1}(0,0)\) radius \({r_1}\, = 9\) \({C_2}(3,4)\) , radius \({r_2}\, = 4\) \({C_1}{C_2}\, = \,|{r_1} - {r_2}|\, = \,5\) \(\therefore \) Circle touches intemally \(\therefore \,|{z_1}\, - \,{z_2}{|_{\min }}\, = \,0\)
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