Question

If $A(2,8)$ 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. $\varphi$

Answer 1

The correct option is D $\varphi$

$A(2,8)$ is an interior point of the given circle.
$\therefore 4+64-4+32-P<0\Rightarrow P>96$ ...(1)
As the given circle neither touches nor intersects the axes.
For $x=0\Rightarrow y^2+4y-P=0$
For $y=0\Rightarrow x^2-2x-P=0$
these equation should not have any real roots.
$\Rightarrow 16+4P<0\&4+4P<0\Rightarrow P<-\mathrm{4.....}\left(2\right)$

From Eq.$\left(1\right)$ and Eq.$\left(2\right),$ $P=\varphi$