Question

If $P(2,8)$ is an interior point of a circle $x^2+y^2-2x+4y-p=0$ which neither touches nor intersects the axes, then set for $p$ is -

• A. p < - 1
• B. p < - 4
• C. p > 96
• D.

Answer 1

The correct option is D

Since the point P is interior to the circle, $S_1$ < 0
$=\left(2^2\right)+\left(8^2\right)-2.\left(2\right)+4\left(8\right)-p<0$
$=96-p<0$
$=p>96$
Also given that the circle doesn't touches any of the axes.
So, $g^2-c$ < 0
$f^2-c$ < 0
$g^2-c$ < 0
$=1+p<0$
$=p<-1$
Also,
$f^2-c$ < 0
$=4+p<0$
$=p<-4$
Since $p<-4$ and $p>96$ is not possible, the set for p will be a null set meaning it'll not have any element in it.