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 the set of all possible values of $p$ is the interval :

• A. $(0,25)$
• B. $(25,39)$
• C. $(9,25)$
• D. $(25,29)$

Answer 1

The correct option is D $(25,29)$

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

$(x-3{)}^2+(y-5{)}^2-9-25+p=0$

$(x-3{)}^2+(y-5{)}^2=34-p$

Now

$34-p>0$

$p<34$ ...(i)

Now point $(1,4)$ lies inside the circle.

Hence

$1+16-6-40+p<0$
Or

$11-40+p<0$

$-29+p<0$

$p<29$ ...(ii)

From i and ii

$p<29$

Now it does not touch or intersect co-ordinate axes.

At $x=0$

$y^2-10y+p>0$ since it will lie outside the circle.

Hence

$b^2-4ac<0$

Or

$100-4p<0$

Or

$p>25$

Hence

$p\in (25,29)$.