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  • The point (1,4) lies inside the circle <img alt="\mathrm{x^2+y^2-6 x-10 y+p=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E2&plus;y%5E2-6%20x-10%20y&plus;p%3D0%7D"

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 axis, then
 

Option: 1

\mathrm{ 0<\mathrm{p}<29}


Option: 2

\mathrm{ 25<\mathrm{p}<29}


Option: 3

\mathrm{ 9<\mathrm{p}<25}


Option: 4

\mathrm{ 9<\mathrm{p}<29 }


Answers (1)

best_answer

Since the circle does not tough 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 \mathrm{\sqrt{9+25-p}}

\mathrm{3>\sqrt{9+25-p} \Rightarrow p>25}\quad \quad \dots(1)

So that \mathrm{5>\sqrt{9+25-\mathrm{p}} \Rightarrow \mathrm{p}>9}\quad \quad \dots(2)

And the point (1,4) lies inside the circle \mathrm{\Rightarrow 1+16-6-10 \times 4+\mathrm{p}<0 \Rightarrow \mathrm{p}<29 }

From (1), (2), (3) we get \mathrm{25<p<29.}

Posted by

himanshu.meshram

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