Question

If $k{x}^2+k{y}^2-4x-6y-2k=0$ represents a real circle, then the set of values of $k$ is

• A. $(0,\infty )$
• B. $(-\infty ,0)$
• C. $R$
• D. $R-\left\{0\right\}$

Answer 1

The correct option is D $R-\left\{0\right\}$

$k{x}^2+k{y}^2-4x-6y-2k=0$
Clearly, $k\ne 0$
Hence, the above equation can be rewritten as,
$x^2+y^2-\frac{4x}{k}-\frac{6y}{k}-2=0$
$\Rightarrow g=-\frac{2}{k},f=-\frac{3}{k},c=-2$

For real circle,
Radius of the circle, $r\ge 0$
$\Rightarrow g^2+f^2-c\ge 0\Rightarrow {(-\frac{2}{k})}^2+{(-\frac{3}{k})}^2+2\ge 0\Rightarrow \frac{4}{k^2}+\frac{9}{k^2}+2\ge 0\Rightarrow 13+2k^2\ge 0$
which is true for all real values of $k.$

Hence, $k\in R-\left\{0\right\}$