JEE Main 2014 · 9 Apr · Shift 1 · Mathematics Q81

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MathematicsSingle correct+4 / -1Question 81 of 90

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 :

(0,25)

(25,39)

(9,25)

(25,29)

Answer and worked solution

Official answer: D

The centre is $(3,5)$ and $r^2=34-p$ .

Not meeting the axes: $r<3$ , so $p>25$ .
$(1,4)$ inside: $1+16-6-40+p<0$ , so $p<29$ .

$p\in(25,29)$ .

Topics:Equation of a circle

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JEE Main 2014 · 9 Apr · Shift 1, Mathematics question 81