The circle ${\mathrm{x}}^2+{\mathrm{y}}^2-6\mathrm{x}-10\mathrm{y}+\mathrm{\lambda }=0$ does not touch or intersect the coordinate axes and the point (1,4) is inside the circle. Then the number of integral values of $\mathrm{\lambda }$ are_______

Let $\mathrm{S}\equiv {\mathrm{x}}^2+{\mathrm{y}}^2-6\mathrm{x}-10\mathrm{y}+\mathrm{\lambda }=0$
$\therefore$Point (1, 4) is inside the circle, then S₁ < 0

$\Rightarrow \begin{array}{c}1+16-6-40+\mathrm{\lambda }<0\\ \mathrm{\lambda }<29⟶\text{ (i) }\end{array}$
Centre and radius of the circle are (3,5) and $\sqrt{(34-\mathrm{\lambda })}$, respectively.
$\because$Circle does not touch or intersect the coordinates axes.
$\therefore 5>\mathrm{r}\text{ and }3>\mathrm{r}$
or $5>\sqrt{(34-\mathrm{\lambda })}\text{ and }3>\sqrt{(34-\mathrm{\lambda })}$
$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}25>34-\mathrm{\lambda }\text{ and }9>34-\mathrm{\lambda }\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{\lambda }>9\text{ and }\mathrm{\lambda }>25\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{\lambda }>25⟶\text{ (ii) }\end{array}$
Also   $34-\mathrm{\lambda }>0$
$\therefore \mathrm{\lambda }<34\to \left(\mathrm{iii}\right)$
From Eqs. (i), (ii) and (iii), we get $25<\mathrm{\lambda }<29$
$\mathrm{\lambda }=$ 26,27,28 total 3 values