A) \[0<p<29\]
B) \[25<p<35\]
C) \[25<p<29\]
D) None of these
Correct Answer: C
Solution :
Given, circle is\[{{x}^{2}}+{{y}^{2}}-6x-10y+p=0\] ...(i) its centre is (3, 5) and radius, \[r=\sqrt{9+25-p}\] \[=\sqrt{34-p}\] Since the circle neither touches the axes nor intersect \[\therefore \] \[x-\]coordinate of centre\[>r\] y-coordinate of centre\[>r\] and \[S(1,4)<0\] \[\Rightarrow \] \[-3>\sqrt{34-p}\]and \[5>\sqrt{34-p}\] On squaring, we get \[9>34-p\] and\[25>34-p\] \[\Rightarrow \] \[p>34-9\]and \[p>34-25\] \[\Rightarrow \] \[p>25\]and \[p>9\] ...(ii) Now, put\[x=1,\text{ }y=4\]in Eq. (i), we get \[1+16-6-40+p<0\] \[\Rightarrow \] \[p<29\] ...(ii) Hence, from Eqs. (ii) and (iii), we get \[25<p<29\]and \[9<p<29\]
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