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  • If P(2,8) is an interior point of a circle  <img alt="\mathrm{x^2+y^2-2 x+4 y-p=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E2&plus;y%5E2-2%20x&plus;4%

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

the axes, then set for p is -

Option: 1

\mathrm{\text { p }<-1}


Option: 2

\mathrm{\text { p }<-4}


Option: 3

\mathrm{\text { p }> 96}


Option: 4

\mathrm{\phi}


Answers (1)

best_answer

For internal point  \mathrm{\mathrm{p}(2,8) 4+64-4+32-\mathrm{p}<0 \Rightarrow \mathrm{p}>96}  and X intercept \mathrm{=2 \sqrt{1+p}}  therefore

\mathrm{1+\mathrm{p}<0}

\mathrm{\Rightarrow \quad \mathrm{p}<-1 }  and Y intercept \mathrm{=2 \sqrt{4+\mathrm{p}} \Rightarrow \mathrm{p}<-4}.

Hence (d) is correct answer.

Posted by

jitender.kumar

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