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JEE Mains · Maths · STD 11 - 4.1 complex nubers

Let the lines \((2-i) z=(2+i) \bar{z}\) and \((2+i) z+(i-2) \bar{z}-4 i=0,\) (here \(\left.i^{2}=-1\right)\) be normal to a circle \(C\). If the line \(iz +\overline{ z }+1+ i =0\) is tangent to this circle \(C\), then its radius is

  1. A \(\frac{3}{\sqrt{2}}\)
  2. B \(\frac{1}{2 \sqrt{2}}\)
  3. C \(3 \sqrt{2}\)
  4. D \(\frac{3}{2 \sqrt{2}}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{3}{2 \sqrt{2}}\)

Step-by-step Solution

Detailed explanation

\((i)\) \((2- i ) z =(2+ i ) \overline{ z }\) \(y=\frac{x}{2}\) \((ii)\) \((2+ i ) z +( i -2) \overline{ z }-4 i =0\) \(x+2 y=2\) \((iii)\) \(iz +\overline{ z }+1+ i =0\) \(Eq ^{ n }\) of tangent \(x - y +1=0\) Solving \((i)\) and \((ii)\) \(x=1, y=\frac{1}{2}\) Now,…
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