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

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(D) The equation of the circle is $x^2 + y^2 - 6x - 10y + p = 0$. Completing the square,we get $(x - 3)^2 + (y - 5)^2 = 34 - p$. The center is $(3, 5)

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$(25, 29)$

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(D) The equation of the circle is $x^2 + y^2 - 6x - 10y + p = 0$. Completing the square,we get $(x - 3)^2 + (y - 5)^2 = 34 - p$. The center is $(3, 5)$ and the radius $r = \sqrt{34 - p}$. For the circle to not touch or intersect the $x$-axis,the radius must be less than the $y$-coordinate of the center: $r  9$. For the circle to not touch or intersect the $y$-axis,the radius must be less than the $x$-coordinate of the center: $r  25$. If the point $(1, 4)$ lies inside the circle,the distance from the center $(3, 5)$ to $(1, 4)$ must be less than the radius: $\sqrt{(3 - 1)^2 + (5 - 4)^2} Combining all conditions,$p > 25$ and $p < 29$,so $p \in (25, 29)$.

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

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