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  • Stuck here, help me understand: - Co-ordinate geometry - JEE Main-2

 If the point (1, 4) lies inside the circle  x^{2}+y^{2}-6x-10y+p=0 and the circle  does not touch or intersect the coordinate axes, then the set of all possible values of p is the interval :

  • Option 1)

    (0,\: \; 25)

  • Option 2)

    (25,\: \; 39)

  • Option 3)

    (9,\: \; 25)

  • Option 4)

    (25,\: \; 29)

 

Answers (1)

As we learnt in

General form of a circle -

x^{2}+y^{2}+2gx+2fy+c= 0
 

- wherein

centre = \left ( -g,-f \right )

radius = \sqrt{g^{2}+f^{2}-c}

 

 

Given

12 + 42 - 6 - 40 + p < 0

p < 29

also r=\sqrt{3^{2}+5^{2}-p}

r=\sqrt{34-p}

\sqrt{34-p}<3 (since it shouldn't touch y-axis)

\Rightarrow     34 - p < 9

\Rightarrow     p > 25

so, p \epsilon (25, 29)


Option 1)

(0,\: \; 25)

This is incorrect option

Option 2)

(25,\: \; 39)

This is incorrect option

Option 3)

(9,\: \; 25)

This is incorrect option

Option 4)

(25,\: \; 29)

This is correct option

Posted by

Sabhrant Ambastha

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