Question

The following equation $\dfrac{{k{{\left( {x + 1} \right)}^2}}}{3} + \dfrac{{{{\left( {y + 2} \right)}^2}}}{{4k}} = 1$ represent a circle. Find the value of $k$.

Answer 1

Convert the given equation in the form of standard general equation of second degree $a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$ and then apply the condition for this equation to represent a circle, i.e., Coefficient of ${x^2}$= Coefficient of ${y^2}$ ($a = b$) and Coefficient of $xy = 0$ ($h = 0$).

Complete step by step solution:
Given equation is $\dfrac{{k{{\left( {x + 1} \right)}^2}}}{3} + \dfrac{{{{\left( {y + 2} \right)}^2}}}{{4k}} = 1$.
Solve the given equation by taking LCM of denominator.
$\Rightarrow \dfrac{{4k \cdot k{{\left( {x + 1} \right)}^2} + 3{{\left( {y + 2} \right)}^2}}}{{3 \times 4k}} = 1$
$\Rightarrow \dfrac{{4{k^2}\left( {{x^2} + 1 + 2x} \right) + 3\left( {{y^2} + 4 + 4y} \right)}}{{12k}} = 1$
$\Rightarrow 4{k^2}\left( {{x^2} + 1 + 2x} \right) + 3\left( {{y^2} + 4 + 4y} \right) = 12k$
$\Rightarrow 4{k^2}{x^2} + 4{k^2} + 8{k^2}x + 3{y^2} + 12 + 12y = 12k$
$\Rightarrow 4{k^2}{x^2} + 3{y^2} + 8{k^2}x + 12y + 4{k^2} - 12k + 12 = 0$ …. (1)
We have given that the above equation (1) represents a circle.
Condition for the general equation of second degree in $x$ and $y$viz., $a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$ to represent a circle is: $a = b$and $h = 0$.
$a = b$ implies that coefficient of ${x^2}$= coefficient of ${y^2}$
$\therefore 4{k^2} = 3$ [from(1)]
$\Rightarrow {k^2} = \dfrac{3}{4}$
On taking square root,
$\Rightarrow k = \sqrt {\dfrac{3}{4}}$
$\Rightarrow k = \dfrac{{\sqrt 3 }}{2}$

Hence the value of $k$ should be $\dfrac{{\sqrt 3 }}{2}$ so that the given equation represents a circle.

Note:
The general equation of second degree in $x$ and $y$viz., $a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$, represent a circle if $a = b$ and $h = 0$. Therefore, the general equation of circle is ${x^2} + {y^2} + 2gx + 2fy + c = 0$, whose centre is $\left( { - g, - f} \right)$ and radius is $\sqrt {{g^2} + {f^2} - c}$.