Question

The equation ${\text{k}}\left( {{{\text{x}}^2} + {{\text{y}}^2}} \right) - {\text{x - y + k = 0}}$ represents a real circle, if
${\text{A}}{\text{. k < }}\sqrt 2 \\\ {\text{B}}{\text{. k > }}\sqrt 2 \\\ {\text{C}}{\text{.|k| < }}\dfrac{1}{{\sqrt 2 }} \\\ {\text{D}}{\text{. 0 < |K| }} \leqslant {\text{ }}\dfrac{1}{{\sqrt 2 }} \\\$

Answer 1

Hint: To check if the equation represents a circle, we transform the given circle equation into the general form of a circle. We then use the condition that represents a real circle to verify.

Complete step-by-step answer:
We know, for an equation ${{\text{x}}^2} + {{\text{y}}^2} + 2{\text{gx + 2fy + c = 0}}$ to represent a real circle, the condition is $\sqrt {\left( {{{\text{g}}^2} + {{\text{f}}^2} - {\text{c}}} \right)} > 0$.

Given equation ${\text{k}}\left( {{{\text{x}}^2} + {{\text{y}}^2}} \right) - {\text{x - y + k = 0}}$
We divide the equation by k and compare it with the circle equation we get,
${{\text{x}}^2} + {{\text{y}}^2} - \dfrac{1}{{\text{k}}}{\text{x - }}\dfrac{1}{{\text{k}}}{\text{y + 1 = 0}}$
$\Rightarrow {\text{g = - }}\dfrac{1}{{2{\text{k}}}}{\text{ , f = - }}\dfrac{1}{{2{\text{k}}}}{\text{ , c = 1}}$

To represent a circle, $\sqrt {\left( {{{\text{g}}^2} + {{\text{f}}^2} - {\text{c}}} \right)} > 0$ must hold true,

$\Rightarrow \sqrt {\left( {{{\left( { - \dfrac{1}{{2{\text{k}}}}} \right)}^2} + {{\left( { - \dfrac{1}{{2{\text{k}}}}} \right)}^2} - 1} \right)} > 0 \\\ \Rightarrow \sqrt {\dfrac{1}{{4{{\text{k}}^2}}} + \dfrac{1}{{4{{\text{k}}^2}}} - 1} {\text{ > 0}} \\\ \Rightarrow \sqrt {\dfrac{1}{{2{{\text{k}}^2}}} - 1} {\text{ > 0}} \\\ \Rightarrow \dfrac{1}{{2{{\text{k}}^2}}}{\text{ > 1}} \\\ \Rightarrow {\text{1 > 2}}{{\text{k}}^2} \\\ \Rightarrow \dfrac{1}{2} > {{\text{k}}^2} \\\ \Rightarrow |{\text{k| < }}\dfrac{1}{{\sqrt 2 }} \\\$

Hence, for the equation ${\text{k}}\left( {{{\text{x}}^2} + {{\text{y}}^2}} \right) - {\text{x - y + k = 0}}$ represents a real circle if ${\text{|k| < }}\dfrac{1}{{\sqrt 2 }}$.
Option C is the right answer.

Note: In order to solve such types of questions the key is to have adequate knowledge of the general equation and conditions for a circle to represent a real circle. And then we modify the given circle into the form of a general equation of a circle and use the required condition, we have to be very careful while comparing the given equation with the general form of circle to obtain the values of g and f.