The equation of given circle is x2+y2−6x−10y+p=0 ⇒(x−3)2+(y−5)2=34−p ∴ The point (1,4) lies inside the circle, so 1+16−6−40+p<0 ⇒p<29.........(i) ∴ The circle neither intersects nor touches the coordinate axis, then r=√34−p<3⇒34−p<9 ⇒p>25 ...(ii) and r=√34−p<5 ⇒34−p<25 ⇒p>9.....(iii) from in equalities Eqs. (i), (ii) and (iii), we get $25