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  • If  is an interior point of a circle <img alt="\mathrm{x}^{2}+\mathrm{y}^{2}-2 \

If  \mathrm{P}(2,8) is an interior point of a circle \mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{x}+4 \mathrm{y}-\mathrm{p}=0 which neither touches nor intersects 2 \sqrt{1+p}  the axes, then set for \mathrm{p} is

Option: 1

\mathrm{p}<-1


Option: 2

\quad \mathrm{p}<-4


Option: 3

\mathrm{p}>96


Option: 4

\phi


Answers (1)

best_answer

For internal point \mathrm{p}(2,8) 4+64-4+32-\mathrm{p}<0 \mathrm{p}>96$ and $\mathrm{x} intercept = therefore

1+\mathrm{p}<0
\mathrm{p}<-1 and \mathrm{y} intercept =2 \sqrt{4+p} \mathrm{p}<-4

Hence (d) is the correct answer.

Posted by

Pankaj Sanodiya

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