If the point (1,4) lies inside the circle x^2 | 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,the radiu

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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,the radius squared must be positive: $34-p > 0 \Rightarrow p Since the circle does not touch or intersect the coordinate axes,the distance from the center $(3,5)$ to the axes must be greater than the radius $r = \sqrt{34-p}$. For the $x$-axis,the distance is $|y_c| = 5$. Thus,$r  9$ ... $(ii)$. For the $y$-axis,the distance is $|x_c| = 3$. Thus,$r  25$ ... $(iii)$. Since the point $(1,4)$ lies inside the circle,substituting it into the circle equation must yield a negative value: $1^2+4^2-6(1)-10(4)+p Combining inequalities $(i), (ii), (iii),$ and $(iv)$,we get $25 Therefore,option $(b)$ is correct.

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

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