TS EAMCET · Maths · Complex Number
Let \(a=1+i\) and \(z=x+i y\). If the curve \(z \bar{z}+a z+\bar{a} \bar{z}-4=0\) is cut by the straight line \((z+\bar{z})-i(z-\bar{z})+2=0\) at two points \(A\) and \(B\), then the equation of the circle passing through the origin, \(A\) and \(B\) is
- A \(x^2+y^2+3 x-4 y=0\)
- B \(x^2+y^2+x+y=0\)
- C \(x^2+y^2+6 x+2 y=0\)
- D \(x^2+y^2-7 x-12 y=0\)
Answer & Solution
Correct Answer
(C) \(x^2+y^2+6 x+2 y=0\)
Step-by-step Solution
Detailed explanation
Circle \(z \bar{z}+a z+\bar{a} \bar{z}-4=0 \quad\left[\because z \bar{z}=|z|^2\right]\) \(\Rightarrow|z|^2+a z+(\overline{a z})-4=0 \quad\left[\because \bar{z}_1 \bar{z}_2=\overline{z_1 z_2}\right]\)…
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