AP EAMCET Engineering 21 Apr 2019 Shift 2 Paper

It is given that
z=x+iy,x,y∈R
And
‌

2z+i

z−2

‌‌=‌

2(x+iy)+i

(x−2)+iy

Re(‌

2z+i

z−2

)‌‌=‌

2x+i(2y+1)

(x−2)+iy

×‌

(x−2)−iy

(x−2)−iy

0‌‌=2x(x−2)+y(2y+1)=0
0‌‌=x2+y2−2x+‌

y

2

‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(I)
And
‌

z+i

z−i

=‌

x+i(y+1)

(x+1)+iy

×‌

(x+1)−iy

(x+1)−iy

Now
arg(‌

z+i

z−i

)=Tan−1[‌

(x+1)(y+1)−xy

x(x+1)+y(y+1)

]=‌

π

2

x2+y2x+x+y=0‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(II)
Equation of common chord of circles (i) and (ii) is
S1−S2=0
−3x−‌

y

2

=0
y+6x=0‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(III)
From equation (II) and (III), the intersection of the circles is,
x2+36x2+x−6x‌‌=0
x‌‌=0,‌

5

37

So the values of y are 0,−‌

30

37

This implies the point of intersection of the curves C1 and C2
other than the origin is (‌

5

37

,−‌

30

37

)