Question

If the point $(1,4)$ lies inside the circle $x^2+y^2-6x-10y+p=0$ and the circle does not touch or intersect the coordinate axes, then

• A. $0<p<34$
• B. $25<p<29$
• C. $9<p<25$
• D. $9<p<29$

Answer 1

Answer: (B)

$25<p<29$

Since the circle does not touch or intersect the coordinates axes,
the
absolute values of $x$ and $y$ coordinates of the centre are greater
than the radius of the circle.
Coordinates of the centre of the circle
are $(3,5)$ and the radius is .$\sqrt{9+25-p}$
$\Rightarrow 3>\sqrt{9+25-p}\Rightarrow P>25...\left(1\right)$
and $5>\sqrt{9+25-p}\Rightarrow P>9....\left(2\right)$
and the point $(1,4)$ lies inside the
circle $\Rightarrow 1+16-6-10\times 4+p<0\Rightarrow p<29....\left(3\right)$
From $\left(1\right),\left(2\right),\left(3\right)$ we get $25<p<29.$