TS EAMCET 5 May 2018 Shift 2 Solved Paper

Let the point P(h,k),A(1,0) and B(x1,y1) such that AP=3AB.
So,
AB+BP=3AB
BP=2AB

AB

BP

=

1

2

Therefore B divides AP in the ration 1: 2 internally.
x1=

h+2

3

y1=

k

3

Now the point (x1,y1) lies on the circle,
x2+y2−2x+2y+1=0
(

h+2

3

)2+(

k

3

)2−2(

h+2

3

) +2(

k

3

)+1=0
h2+4h+4+k2−6h −12+6k+9=0
h2+k2−2h+6k+1=0
Therefore, the locus of P is,
x2+y2−2x+6y+1=0