TS EAMCET 10-Sep-2020 Shift 1 Solved Paper

Let the equation of circle S is
S=x2+y2+2gx+2fy+c=0
C1=x2+y2−8x−2y+16=0
C2=x2+y2−4x−4y−1=0
mathcal‌S is orthogonally to C1 and C2
∴‌‌−8g−2f=C+16...(i)
−4g−4f=C−1.....(ii)
From Eqs. (i) and (ii),
−4g+2f=17...(iii)
Common chord of S and C1=

Given common chord is
2x+13y−15=0...(v)
Eqs. (iv) and (v) are coincide
∴‌

2g+8

2

=‌

2f+2

13

=‌

c−16

−15

⇒13g−2f=−50...(vi)
Solving Eqs. (iii) and (vi), we get
g=‌

11

3

f=‌

−7

6

∴ Centre of circle, S=(‌

−11

3

,‌

7

6

)