The circle x^2+y^2-6x-10y+p=0 neither interse | vedclass

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(B) The equation of the given circle is $x^2+y^2-6x-10y+p=0$. Completing the square,we get $(x-3)^2+(y-5)^2 = 34-p$. For the circle to exist,$34-p > 0

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$25 < p < 29$

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(B) The equation of the given circle is $x^2+y^2-6x-10y+p=0$. Completing the square,we get $(x-3)^2+(y-5)^2 = 34-p$. For the circle to exist,$34-p > 0$,so $p The center is $(3,5)$ and the radius is $r = \sqrt{34-p}$. Since the point $(1,4)$ lies inside the circle,the power of the point must be negative: $1^2+4^2-6(1)-10(4)+p Since the circle does not intersect or touch the coordinate axes,the distance from the center $(3,5)$ to the axes must be greater than the radius $r$. Distance to $y$-axis is $|3| = 3$,so $r  25$ (ii). Distance to $x$-axis is $|5| = 5$,so $r  9$ (iii). Combining $(i)$,(ii),and (iii),we get $25 < p < 29$.

The circle $x^2+y^2-6x-10y+p=0$ neither intersects nor touches the coordinate axes and the point $(1,4)$ lies inside the circle. Then the range of possible values of $p$ is

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