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:

The equation of the circle is $$x^2 + y^2 - 6x - 10y + P = 0$$

$$(x - 3)^2 + (y - 5)^2 = \left(\sqrt{34 - P}\right)^2 \qquad \text{......(i)}$$

Centre = $$(3, 5)$$ and radius $$r = \sqrt{34 - P}$$

If circle does not touch or intersect the $$x\text{ - axis}$$ then radius $$<$$ $$y\text{ - coordinate}$$ of centre $$C$$ or $$\sqrt{34 - P} < 5$$

$$\implies 34 - P < 25 \implies P > 9 \qquad \text{......(ii)}$$

Also if the circle does not touch or intersect $$y\text{ - axis}$$ the radius $$<$$ $$x\text{ - coordinate}$$ of centre $$C$$.

$$\text{or } \sqrt{34 - P} < 3 \implies 34 - P < 9 \implies P > 25 \qquad \text{.......(iii)}$$

If the point $$(1, 4)$$ is inside the circle, then its distance from centre $$C < r$$.

$$\text{or } \sqrt{(3 - 1)^2 + (5 - 4)^2} < \sqrt{34 - P} \implies 5 < 34 - P$$

$$\implies P < 29 \qquad \text{.........(iv)}$$

Now all the conditions $$\text{(ii)}$$, $$\text{(iii)}$$ and $$\text{(iv)}$$ are satisfied if $$25 < P < 29$$ which is required value of $$P$$.