The circle \(x^2+y^2-6 x-10 y+p=0\) neither intersects nor touch the coordinate axes and the point \((1,4)\)…
The circle \(x^2+y^2-6 x-10 y+p=0\) neither intersects nor touch the coordinate axes and the point \((1,4)\) lies inside the circle. Then the range of possible values of ' \(p\) ' is
\(23 < p < 25\)
\(25 < p < 29\)
\(21 < p < 23\)
\(12 < p < 21\)
Solution
The equation of given circle is
\(\begin{array}{rlrl}
x^2+y^2-6 x-10 y+p & =0 \\
\Rightarrow & \quad(x-3)^2+(y-5)^2 & =34-p
\end{array}\)
\(\therefore\) The point \((1,4)\) lies inside the circle, so
\(\begin{array}{ll}
& 1+16-6-40+p < 0 \\
\therefore \quad & p < 29 \quad \ldots (i)
\end{array}\)
\(\therefore\) The circle neither intersects nor touches the coordinate axis, then
\(\begin{array}{lll}
& r=\sqrt{34-p} < 3 \Rightarrow 34-p < 9 \\
\Rightarrow & p > 25 \quad \ldots (ii) \\
\text {and } & r=\sqrt{34-p} < 5 \\
\Rightarrow & 34-p < 25 \\
\Rightarrow & p > 9 \quad \ldots (iii)
\end{array}\)
from in equalities Eqs. (i), (ii) and (iii), we get \(25 < p < 29\)
Hence, option (b) is correct.