question_answer

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:   [JEE Main Online Paper ( Held On 09 Apirl  2014  )

A) \[(0, 25)\]                 

B) \[(25, 39)\]

C) \[(9, 25)\]                 

D) \[(25, 29)\]

Correct Answer: D

Solution :

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