TS EAMCET 4 May 2018 Shift 1 Solved Paper

The equation of the required circle is,
x2+y2+2gx+2fy+c=0
The circle cuts, x2+y2−2x+6y=0, x2+y2−4x−2y+6=0 and, x2+y2−12x+2y+3=0 orthogonally.
Then,
−2g+6f=c
−4g−2f=c+6
−12g+2f=c+3
Solve the above three equations for f,g,c
g=0
f=−

3

4

c=−

9

2

The equation of the required circle is
x2+y2−

3

2

y−

9

2

=0
Therefore, the equation of tangent at point (0,3) is,
3y−

3

4

(y+3)−

9

2

=0
y=3 .