Range of $k$ in $x^2+y^2-6x-10y+k=0$ 
The circle $x^2+y^2-6x-10y+k=0$ does not touch or intersect the x-axis and the point $(1,4)$ lies inside the circle, then find the range of $k$

$$
C(3,5)\;\&\;r=\sqrt{34-k}
$$
Attempt 1
$$
d=\frac{|ax+by+c|}{\sqrt{a^2+b^2}}=|3|>r=\sqrt{34-k}\implies k>25
$$
Attempt 2
$$
y=0\implies x^2-6x+k=0\implies\Delta=36-4k<0
\implies k>9
$$
The solution given in my reference is $9<k<29$. The upper limit is fine, but which case is missing in attempt 1 that I am not getting the lowest limit for $k$ there ?

The distance from center $C(3,5)$ to the $x-$ axis is $d=5$ .
It must be greater than the radius.
thus
$$5 >\sqrt{34-k}$$
the distance from the center to the point $(1,4)$ is
$D=\sqrt{(3-1)^2+(5-4)^2}=\sqrt{5}$ 
It must be smaller than the radius. thus
$$\sqrt{5}<\sqrt{34-k}$$
So
$$5<34-k<25$$
or
$$9<k<29.$$
