AP EAMCET Engineering 22 Apr 2019 Shift 1 Paper

Simplifying the given expression, we get
arg (

z−1

z−3i

) =

π

2

This implies,
arg (

z−1

z−3i

) =

π

2

arg(z−1)−arg(z−3i)=

π

2

arg[(x−1)+iy]−arg[x+(y−3)i] =

π

2

tan−1

y

x−1

−tan−1

y−3

x

=

π

2

Further simplify the above,
tan−1[

y x−1 − y−3 x

1+ y x−1 y−3 x

]=

π

2

xy−(x−1)(y−3)

x(x−1)+y(y−3)

=tan‌

π

2

xy−(x−1)(y−3)

x(x−1)+y(y−3)

=

1

0

x(x−1)+y(y−3)=0
x2+y2−x−3y=0
The above equation further can be written as,
(x−

1

2

)2+(y−

3

2

)2=

1

4

+

9

4

(x−

1

2

)2+(y−

3

2

)2=(

√10

2

)2
The above represents the circle with center (

1

2

,

3

2

) and radius(

√10

2

)2
Thus,
z∊C:(3−i)z+(3+i)

−

z

−6>0, |z−

1+3i

2

|=(x−

1

2

)2+(y−

3

2

)2 =

√10

2