TS EAMCET 10-Sep-2020 Shift 2 Solved Paper

x2+y2+2gx+2fy+c=0 cuts orthogonally
the circle x2+y2−4x−6y+11=0 and
x2+y2−10x−4y+21=0
∴‌‌−4g−6f=c+11 . . . (i)
and ‌‌−10g−4f=c+21 . . . (ii)
Centre of circle x2+y2+2gx+2fy+c=0 is lie in bisector of angle betweenthe positive coordinate axis
∴ g=f . . . (iii)
⋅s From Eqs. (i) and (iii), we get
−4f−6f=c+11
−10f=c+11 . . . (iv)
From Eqs. (ii) and (iii), we get
−14f=c+21 . . . (v)
From Eqs. (iv) and (v), we get
g=f=‌

−5

2

,c=14
∴ 2g+2f+c−5−5+14=4