Let (x1,y1) lie on the line 2kx+3y−1=0 The equation of the chord to the circle is, x2+y2−2x−4y−4=0 And, xx1+yy1−
2(x+x1)
2
−
4(y+y1)
2
−4=0 x(x1−1)+y(y1−2)−x1−2y1−4=0 The lines 2x+y+5=0 and 2kx+3y+1=0 are conjugate lines. So, 2x+y+5=0 And x(x1−1)+y(y1−2)−x1−2y1−4=0 coincides. Therefore,
2
x1−1
=
1
y1−2
=
−5
x1+2y1+4
=
1
λ
So, x1=2λ+1 y1=λ+2 x1+y1+4=−5λ On solving the above three equations. λ=−1 x1=−1 y1=1 Thus, the value of k is obtained as, 2kx+3y−1=0 −2k+3−1=0 k=1