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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;

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 or intersects the  axes, then p is given by

Option: 1

p < -1


Option: 2

p < -2


Option: 3

 p > 92


Option: 4

\varphi


Answers (1)

best_answer

For internal point P(2, 8)

4 + 64 – 8 +32 – p < 0 

⇒  p > 92 

and  x intercept \mathrm{=2 \sqrt{1+p} \text { therefore } 1+p<0}

\mathrm{\Rightarrow p<-1 \text { and } y \text { intercept }=2 \sqrt{2+p} \Rightarrow p<-2}

 

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Sayak

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