JEE Main 2021 — Mathematics Algebra

Given that

$$\mathrm{arg}(\frac{z-2}{z+2})=\frac{\pi }{4}$$arg(z+2z−2​)=4π​

Let $z=x+iy$z=x+iy

$$\Rightarrow \mathrm{arg}(\frac{x+iy-2}{x+iy+2})=\frac{\pi }{4}$$⇒arg(x+iy+2x+iy−2​)=4π​

$$\Rightarrow \mathrm{arg}((\frac{x+iy-2}{x+iy+2})(\frac{x+2-iy}{x+2-iy}))=\frac{\pi }{4}$$⇒arg((x+iy+2x+iy−2​)(x+2−iyx+2−iy​))=4π​

$$\Rightarrow \mathrm{arg}(\frac{{x}^{2}+{y}^{2}-4+4iy}{(x+2{)}^{2}+{y}^{2}})=\frac{\pi }{4}$$⇒arg((x+2)2+y2x2+y2−4+4iy​)=4π​

$$\Rightarrow \frac{{x}^{2}+{y}^{2}-4+4iy}{(x+2{)}^{2}+{y}^{2}}=\mathrm{tan}(\frac{\pi }{4})$$⇒(x+2)2+y2x2+y2−4+4iy​=tan(4π​)

$$\Rightarrow \frac{{x}^{2}+{y}^{2}-4}{4y}=1$$⇒4yx2+y2−4​=1

${x}^{2}+(y-2{)}^{2}=8$x2+(y−2)2=8 ……….(1)

Equation $(1)$(1) represents circle, whose centre $(0,2)$(0,2), radius $2\sqrt{2}$22​.

Minimum distance $d=|AC-r|$d=∣AC−r∣

Given point $A(9\sqrt{2},2)$A(92​,2)

$$=|AC-r|=\mid (9\sqrt{2}-2\sqrt{2})=7\sqrt{2}$$=∣AC−r∣=∣(92​−22​)=72​

$\Rightarrow$⇒ Square of distance $={d}^{2}=(7\sqrt{2}{)}^{2}=98$=d2=(72​)2=98