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यदि बिंदु $(1,4)$ वृत्त $x^{2}+y^{2}-6 x-10 y+p=0$ के अन्त: भाग में स्थित है तथा वृत्त, निर्देशांक अक्षों को न तो स्पर्श करता है, और न ही काटता है, तो $p$ के सभी संभव मानों का समुच्चय निम्न अतंराल है
$(0,25)$
$(25, 39)$
$(9, 25)$
$(25, 29)$
Solution

The equation of circle is
${x^2} + {y^2} – 6x – 10y + P = 0\,\,\,\,\,……\left( i \right)$
${\left( {x – 3} \right)^2} + {\left( {y – 5} \right)^2} = {\left( {\sqrt {34 – P} } \right)^2}$
center $(3,5)$ and radius $'r' = {\left( {\sqrt {34 – P} } \right)^2}$
If cicrle does not touch or intersect the $x$-axis then radium $x < y$ -coordiante of center $C$
or $\sqrt {34 – P} < 5$
$ \Rightarrow 34 – P < 5$
$ \Rightarrow P > 9\,\,\,\,\,\,……\left( {ii} \right)$
Also if the circle does not touch or intersect $X$-axisthe radius $C$
or $\sqrt {34 – P} < 3 \Rightarrow 34 – P < 9 \Rightarrow P > 25\,\,\,\,\,\,……\left( {iii} \right)$
If the point $(1,4)$ is inside the circle, then its distance from center $C < r$
or $\sqrt {\left[ {{{\left( {3 – 1} \right)}^2} + {{\left( {5 – 4} \right)}^2}} \right]} < \sqrt {34 – P} $
$ \Rightarrow 5 < 34 – K$
$ \Rightarrow P < 29\,\,\,\,\,\,\,\,\,\,\,\,\,……..\left( {iv} \right)$
Now all the condition $(ii)$, $(ii)$ and $(iv)$ are satisfied if $25 < P < 29$ which is required value of $P$.