Question

If the equation $\dfrac{K{{\left( x+1 \right)}^{2}}}{3}+\dfrac{{{\left( y+2 \right)}^{2}}}{4}=1$ represents a circle, then K is equal to
(a) $\dfrac{3}{4}$
(b) 1
(c) $\dfrac{4}{3}$
(d) 12

Answer 1

Hint: Calculate eccentricity of the circle using the formula that eccentricity of any ellipse of the form $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ is given by $e=\dfrac{\sqrt{{{a}^{2}}-{{b}^{2}}}}{a}$. Equate it to zero to find the value of K.

Complete step-by-step answer:
We have the equation of the curve as $\dfrac{K{{\left( x+1 \right)}^{2}}}{3}+\dfrac{{{\left( y+2 \right)}^{2}}}{4}=1$. If this equation of curve represents a circle, its eccentricity should be equal to zero.
Eccentricity of any curve is a non-negative real number that uniquely characterizes its shape, i.e. , two conic sections are similar if they have the same eccentricity.
Eccentricity of any curve is the constant ratio of distance between any point on a curve and its locus to any point on the curve and its directrix.
The eccentricity of an ellipse (or equivalently a circle) is the distance between its centre and either of its two foci. The eccentricity of any ellipse is strictly less than 1. However, when a circle (with eccentricity equal to zero) is counted as an ellipse, the eccentricity of ellipses is greater than or equal to zero.
We know that eccentricity of any ellipse of the form $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1.....\left( 1 \right)$ is the constant ratio denoted by $e=\dfrac{\sqrt{{{a}^{2}}-{{b}^{2}}}}{a}$.
Hence, for the circle to have eccentricity equal to zero, we must have ${{a}^{2}}={{b}^{2}}$ for the given equation of ellipse.
We will consider the curve $\dfrac{K{{\left( x+1 \right)}^{2}}}{3}+\dfrac{{{\left( y+2 \right)}^{2}}}{4}=1$. By comparing it with the equation of ellipse, we observe that ${{a}^{2}}=\dfrac{3}{K},{{b}^{2}}=4$.
Thus, we have ${{a}^{2}}={{b}^{2}}\Rightarrow \dfrac{3}{K}=4$.
So, by rearranging the terms, we get $K=\dfrac{3}{4}$.
Hence, the eccentricity of the circle must be $\dfrac{3}{4}$, which is option (a).

Note: We can also solve this question by expanding the complete equation of the curve and then comparing it with equation of the form $a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$ to find its eccentricity.