TS EAMCET 5 May 2018 Shift 1 Solved Paper

The equation of normals is expressed as,
x2−3xy−3x+9y=0
(x−3y)(x−3)=0
If, x=3 then,
y=1
Thus, the coordinate of centre is (3,1) .
The equation of the circle is,
x2+y2−6x+6y+17=0
The radius of circle is calculated as,
r1=√32+(−3)2−17
=1
The distance between the centers is calculated as,
r1+r2 =√(3−3)2+(1−(−3))2
r2=3
The required equation of the circle is expressed as,
x2−6x+9+y2−2y +1=9
x2+y2−6x−2y+1=0