Question

If the radius of the circle $x^2+y^2+2\lambda x-2\lambda y+18=0$ cannot exceed $6$, then the number of integral value(s) of $\lambda$ is

Answer 1

Given equation of the circle is
$x^2+y^2+2\lambda x-2\lambda y+18=0$
$g=-\lambda ,f=\lambda ,c=18$
We know that,
$r\le 6$
$\Rightarrow \sqrt{{\lambda }^2+{\lambda }^2-18}\le 6$
For the square root to be defined,
$2{\lambda }^2-18\ge 0\Rightarrow {\lambda }^2\ge 9\Rightarrow \lambda \in (-\infty ,-3]\cup [3,\infty )\cdots \left(1\right)$

Now,
$2{\lambda }^2-18\le 362{\lambda }^2\le 54\Rightarrow {\lambda }^2\le 27\Rightarrow \lambda \in [-3\sqrt{3},3\sqrt{3}]\cdots \left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$, we get
$\lambda \in [-3\sqrt{3},-3]\cup [3,3\sqrt{3}]\Rightarrow \lambda =-5,-4,-3,3,4,5$

Hence, the required number of values of $\lambda$ is $6$.