Question

The number of integral values of $\lambda$ for which the equation $x^2+y^2+\lambda x+(1-\lambda )y+5=0$ is the equation of a circle whose radius cannot exceed 5, is

• A. 14
• B. 18
• C. 16
• D. None of these

Answer 1

The correct option is C

16

According to the question :
$\sqrt{{\left(\frac{-\lambda }{2}\right)}^2+{\left(\frac{\lambda -1}{2}\right)}^2}-5\le 5$

$\Rightarrow {\left(\frac{-\lambda }{2}\right)}^2+{\left(\frac{\lambda -1}{2}\right)}^2\le 30$

${\lambda }^2+(\lambda -1{)}^2\le 120$

$\Rightarrow 2{\lambda }^2-2\lambda -119\le 0$

Using quadratic formula:

$\Rightarrow \lambda =\frac{2\pm \sqrt{2^2-4\left(2\right)(-119)}}{2\left(2\right)}$

$\Rightarrow \lambda =\frac{2\pm \sqrt{956}}{4}$

$\Rightarrow \lambda =\frac{1\pm \sqrt{239}}{2}$

$\Rightarrow \lambda =-7.23,8523-7.23\le \lambda \le 8.23$

$\Rightarrow \lambda =-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8\left(If\lambda \epsilon z\right)$

Thus, the number of integral values of $\lambda$ is 16.