If the point (1, 4) lies inside the circle x ^ 2 + y ^ 2 - 6x + 10y...

{x}^{2} + {y}^{2} - 6x - 10y + p = 0$</p><p>$∴$Centre is \left(3, 5\right)$ and radius is $\sqrt{34 - p}$.

$∵ \left(1, 4\right) lies inside$ the circle.

So,1 + 16 - 6 - 40 + p < 0$</p><p>$⇒ p < 29 . . . \left(1\right)$</p><p>Circle neither touches nor cut the coordinate axes.<br />So, radius of circle$\sqrt{34 - p} < 5$(for not touching$x -$axis )$ and \sqrt{34 - p} < 3$(for not touching$y -$axis)</p><p>$⇒ 9 < p and 25 < p$</p><p>$⇒ p > 25 . . . \left(2\right)

From equations $\left(1\right)\text{\text{ and } }$\left(2\right),

So, $p \in \left(25, 29\right)$