JEE Main 2014 — Circle Question with Solution

$x^2+y^2-6x-10y+p=0$

 $\therefore$ Centre is $\left(3, 5\right)$ and radius is $\sqrt{34-p}$.

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

So, $1+16-6-40+p<0$

$\Rightarrow p<29 ...\left(1\right)$   

Circle neither touches nor cut the coordinate axes.
So, radius of circle $\sqrt{34-p}<5$ (for not touching $x-$ axis ) and $\sqrt{34-p}<3$ (for not touching $y-$axis)

$\Rightarrow 9<\mathrm{p}$ and $25<\mathrm{p}\mathrm{ }$

$\Rightarrow p>25 ...\left(2\right)$

From equations $\left(1\right) \& \left(2\right)$,

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