જો બિંદુ (1, 4) એ વર્તુળ x^2 + y^2-6x - 10y + | vedclass

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Question

The equation of circle is 

${x^2} + {y^2} - 6x - 10y + P = 0\,\,\,\,\,......\left( i \right)$

${\left( {x - 3} \right)^2} + {\left( {y - 5}

Vedclass · Accepted Answer

$(25, 29)$

Vedclass · Answer

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 $P$.

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