10-1.Circle and System of Circles
hard

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

A

$(0,25)$

B

$(25, 39)$

C

$(9, 25)$

D

$(25, 29)$

(JEE MAIN-2014)

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 $P$.

Standard 11
Mathematics

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