Q48 / 160 · TS EAMCET 2020 (11 Sep Shift 2) · Mathematics

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
    $x^2+y^2+x+y=0$
  • Correct answer
    C
    $x^2+y^2+6 x+2 y=0$
  • D
    $x^2+y^2-7 x-12 y=0$

Solution

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 $\Rightarrow \quad x^2+y^2+2 x-2 y-4+4(x+y+1)=0$ $\Rightarrow \quad x^2+y^2+6 x+2 y=0$

Hence, the correct answer is option: 'C'.

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