Question

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

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

Answer 1

Answer: (C)

16

Answer 2

Centre of circle $= \left( - \frac{\lambda}{2}, - \frac{(1 - \lambda)}{2} \right)$ ;

Radius of circle $= \sqrt{\left( \frac{\lambda}{2} \right)^{2} + \left( \frac{1 - \lambda}{2} \right)^{2} - 5} \leq 5$

$\Rightarrow 2\lambda^{2} - 2\lambda - 119 \leq 0$ ,

$\therefore$ $\frac{1 - \sqrt{239}}{2} \leq \lambda \leq \frac{1 + \sqrt{239}}{2} \Rightarrow - 7.2 \leq \lambda \leq 8.2$ (Nearly). $\therefore\lambda = - 7, - 6,$ ................,7, 8.

Hence number of integral values of $\lambda$ is 16.