Question

If point $(2,4)$ is interior to the circle $x^2+y^2-6x-10y+k=0$ and the circle does not cut the axes at any point then $k$ belongs to the interval

• A. $(25,32)$
• B. $(9,32)$
• C. $(32,+\infty )$
• D. none of these

Answer 1

Answer: (A)

$(25,32)$

Since, the point $(2,4)$ is interior to the given circle,
$\therefore 2^2+4^2-6\times 2-10\times 4+k<0$
$\Rightarrow k-32<0\Rightarrow k<32$ ...(1)
Solving $y=0,x^2+y^2-6x-10y+k=0$
we get $x^2-6x+k=0,$ which must have imaginary roots.
$\therefore$ Discriminant $=36-4k<0\Rightarrow k>9$ ...(2)
Again, solving $x=0,x^2+y^2-6x-10y+k=0$,
We get $y^2-10y+k=0$, which must have imaginary roots
$\therefore$ Discriminant $=100-4k<0\Rightarrow k>25$ ...(3)
From (1), (2) and (3) we get, $25<k<32$