The correct option is B (25,32)
Since, the point (2,4) is interior to the given circle,
∴22+42−6×2−10×4+k<0
⇒k−32<0⇒k<32 ...(1)
Solving y=0,x2+y2−6x−10y+k=0
we get x2−6x+k=0, which must have imaginary roots.
∴ Discriminant =36−4k<0⇒k>9 ...(2)
Again, solving x=0,x2+y2−6x−10y+k=0,
We get y2−10y+k=0, which must have imaginary roots
∴ Discriminant =100−4k<0⇒k>25 ...(3)
From (1), (2) and (3) we get, 25<k<32