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Question: Answered & Verified by Expert
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
MathematicsComplex NumberTS EAMCETTS EAMCET 2020 (11 Sep Shift 2)
Options:
  • 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 $x^2+y^2-7 x-12 y=0$
Solution:
1983 Upvotes Verified Answer
The correct answer is: $x^2+y^2+6 x+2 y=0$
Circle $z \bar{z}+a z+\bar{a} \bar{z}-4=0 \quad\left[\because z \bar{z}=|z|^2\right]$
$\Rightarrow|z|^2+a z+(\overline{a z})-4=0 \quad\left[\because \bar{z}_1 \bar{z}_2=\overline{z_1 z_2}\right]$
$\Rightarrow|z|^2+2 \cdot \operatorname{Re}(a z)-4=0 \quad \ldots(\mathrm{i})[z+\bar{z}=2 \operatorname{Re}(z)]$
Now, given $z=x+i y \Rightarrow|z|^2=x^2+y^2$
and $a=1+i$, then $a z=(1+i)(x+i y)$
$\therefore a z=(x-y)+i(x+y) \Rightarrow \operatorname{Re}(a z)=x-y$
$\therefore$ From Eq. (i), put $\operatorname{Re}(a z)=x-y$
and $\quad|z|^2=x^2+y^2$
$\Rightarrow \quad x^2+y^2+2(x-y)-4=0$
$\Rightarrow S: x^2+y^2+2 x-2 y-4=0 \ldots(\mathrm{ii})$
is a circle and given line
$L: z+\bar{z}-i(z-\bar{z})+2=0$
$\Rightarrow \quad L: 2 \operatorname{Re}(z)-i 2 \operatorname{Im}(z)+2=0$
$\Rightarrow \quad L: 2 x-i(2 y i)+2=0$
$\Rightarrow \quad L: x+y+1=0 \ldots(\mathrm{iii})$
Now, equation of circle passing through point of intersection of circle $S$ and line $L$ is
$S+\lambda L=0$
$\left(x^2+y^2+2 x-2 y-4\right)+\lambda(x+y+1)=0 \ldots(\mathrm{iv})$
Pass through $(0,0)$
$\therefore$ Put in Eq. (iv), we get
$-4+\lambda=0 \Rightarrow \lambda=4$
Now, required circle is
$\Rightarrow \quad x^2+y^2+2 x-2 y-4+4(x+y+1)=0$
$\Rightarrow \quad x^2+y^2+6 x+2 y=0$

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