Question

If the circle $x^2+y^2-6x-10y+c=0$ does not touch or intersect the coordinate axes and $(1,4)$ lies inside the circle, then the number of integral values of $c$ is

Answer 1

$x^2+y^2-6x-10y+c=0$
Centre is $(3,5)$ and radius $=\sqrt{3^2+5^2-c}=\sqrt{34-c}$
$34-c>0\Rightarrow c<34\cdots \left(1\right)$

Since, $P(1,4)$ lies inside the circle,
$CP<r$
$\Rightarrow (3-1{)}^2+(5-4{)}^2<34-c$
$\Rightarrow c<29\cdots \left(2\right)$
Also, since the circle does not touch or intersect the coordinate axes, we have
$r<5$ and $r<3$
$\Rightarrow r<3$
$\Rightarrow \sqrt{34-c}<3$
$\Rightarrow c>25\cdots \left(3\right)$

From $\left(1\right),\left(2\right)$ and $\left(3\right)$,
$25<c<29$
Possible integral value of $c$ is $26,27,28$