Question

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

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

Answer 1

The correct option is A

$(25,32)$

Explanation for the correct option:

Step 1. To find the interval in which $\lambda$ belongs

Since point $(2,4)$ is interior to the circle $x^2+y^2–6x–10y+\lambda =0$

$\begin{array}{rl}{2}^2+4^2-6\left(2\right)-10\left(4\right)+\lambda & <0\end{array}$

$\Rightarrow$ $\begin{array}{rl}4+16-12-40+\lambda & <0\end{array}$

$\Rightarrow$ $\begin{array}{rl}-32+\lambda & <0\end{array}$

$\Rightarrow$ $\begin{array}{rl}\lambda & <32\end{array}$ …(1)

Step 2. Solving $y=0,$$x^2+y^2–6x–10y+\lambda =0$, we get

$x^2-6x+\lambda =0$ which must have imaginary roots

$\therefore$Discriminant $=36-4\lambda <0$

$\Rightarrow$ $\lambda >9$ …(2)

Step 3. Solving $x=0,$$x^2+y^2–6x–10y+\lambda =0$, we get

$\therefore$Discriminant $=100-4\lambda <0$

$\Rightarrow$ $\lambda >25$ …(3)

From equation (1),(2) and (3), we get

$25<\lambda <32$.

$\mathbf{\therefore }$ $\mathit{\lambda }\mathbf{\in }\mathbf{(}\mathbf{25}\mathbf{,}\mathbf{32}\mathbf{)}$

Hence, Option ‘A’ is Correct.