If P(2, 8) is an interior point of a circle x | vedclass

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(D) For $P(2, 8)$ to be an interior point of the circle $x^2 + y^2 - 2x + 4y - p = 0$,the power of the point must be negative:$2^2 + 8^2 - 2(2) + 4(8)

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$\phi$

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(D) For $P(2, 8)$ to be an interior point of the circle $x^2 + y^2 - 2x + 4y - p = 0$,the power of the point must be negative:$2^2 + 8^2 - 2(2) + 4(8) - p $4 + 64 - 4 + 32 - p $96 - p  96$  .........$(1)$For the circle not to intersect or touch the $x$-axis,the $x$-intercept must be imaginary:$x^2 - 2x + (1 - p) = 0$ has no real roots $\Rightarrow D_x $(-2)^2 - 4(1 - p) Also,the center is $(1, -2)$ and radius $r = \sqrt{1^2 + (-2)^2 + p} = \sqrt{5 + p}$.For no intersection with $x$-axis,$|y_{center}| > r$ $\Rightarrow |-2| > \sqrt{5 + p}$ $\Rightarrow 4 > 5 + p$ $\Rightarrow p For no intersection with $y$-axis,$|x_{center}| > r$ $\Rightarrow |1| > \sqrt{5 + p}$ $\Rightarrow 1 > 5 + p$ $\Rightarrow p Taking the intersection of $(1), (2),$ and $(3)$,we get $p > 96$ and $p Therefore,the set for $p$ is $\phi$.

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 the set for $p$ is

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