Given equation of circle x2+y2+5kx+2y+k=0 and 2x2+2y2+2kx+3y−1=0 Equation of common chord 8kx+y+2k+1=0 Since, this chord is coincide with ‌4x+5y−k=0 ‌‌
8k
4
=‌
1
5
=‌
2k+1
−k
Solving ‌
8k
4
=‌
1
5
we get k=‌
1
10
Solving 8‌
k
4
=‌
2k+1
−k
⇒2k2+2k+1=0 No solution ∴ No values of k possible