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  • The point lies inside the circle <img alt="\mathrm{x^2+y^2-6 x-10 y+p=0}" src="https://entrancecorner.oncodecog

The point (1,4) lies inside the circle \mathrm{x^2+y^2-6 x-10 y+p=0} which does not touch or intersect the coordinate axes, then
 

Option: 1

\mathrm{0<p<29}

 


Option: 2

\mathrm{25<p<29}
 


Option: 3

\mathrm{9<p<25}
 


Option: 4

\mathrm{9<p<29}


Answers (1)

best_answer

Since the circle does not touch or intersect the coordinates axes, the absolute values of \mathrm{x\: and \: \mathrm{y}} coordinates of the centre are greater than the radius of the circle. Coordinates of the centre of the circle are (3,5) and the radius is \mathrm{\sqrt{9+25-p}}
so that\mathrm{ 3>\sqrt{9+25-p} \quad \Rightarrow p>25 }...............(i)

\mathrm{ 5>\sqrt{9+25-p} \quad \Rightarrow p>9}..................(ii)

\thereforePoint (1,4) lies inside the circle \mathrm{x^2+y^2-6 x-10 y+p=0}

\mathrm{\Rightarrow 1+16-6-10 \times 4+p<0 \Rightarrow p<29}............(iii)

From (i), (ii), (iii) we get \mathrm{25<p<29}

Hence option 2 is correct.




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Rishi

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