Q.
The circle S=0 cuts the circles
C1=x2+y2−8x−2y+16=0 and
C2=x2+y2−4x−4y−1=0 orthogonally. If the common chord of S=0 and C1=0 is 2x+13y−15=0, then the centre of S=0
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
matheal S 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
=(2g+8)x+(2f+2)y+c−16=0….(iv)
Given common chord is 2x+13y−15=0…(v)
Eqs. (iv) and (v) are coincide ∴22g+8=132f+2=−15c−16 ⇒13g−2f=−50… (vi)
Solving Eqs. (iii) and (vi), we get g=311f=6−7 ∴ Centre of circle, S=(3−11,67)