Question

If the equation $k\dfrac{{{{\left( {x + 1} \right)}^2}}}{3} + \dfrac{{{{\left( {y + 2} \right)}^2}}}{4} = 1$ represents the circle, then value of $k$ is
A. $\dfrac{3}{4}$
B. $1$
C. $\dfrac{3}{2}$
D. none of these

Answer 1

Hint: In this question, we are given the equation of the circle i.e., $k\dfrac{{{{\left( {x + 1} \right)}^2}}}{3} + \dfrac{{{{\left( {y + 2} \right)}^2}}}{4} = 1$ and we have to find the value of $k$. Now, to find the value step is to take the value of $k$ equals to the values given in options and put it in the equation of the circle. If the equation will come in the standard form of circle it means the value is correct. Otherwise go for the next option.

Formula Used:
Equation of Circle (Standard form) –
${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$, here $\left( {h,k} \right)$ is the center and $r$ is the radius of the circle

Complete step by step solution:
Given that,
Equation of the circle is $k\dfrac{{{{\left( {x + 1} \right)}^2}}}{3} + \dfrac{{{{\left( {y + 2} \right)}^2}}}{4} = 1 - - - - - \left( 1 \right)$
Standard equation of the circle is ${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2} - - - - - - \left( 2 \right)$
Now, put $k = \dfrac{3}{4}$ in equation (1)
$\dfrac{3}{4} \times \dfrac{{{{\left( {x + 1} \right)}^2}}}{3} + \dfrac{{{{\left( {y + 2} \right)}^2}}}{4} = 1$
$\dfrac{{{{\left( {x + 1} \right)}^2}}}{4} + \dfrac{{{{\left( {y + 2} \right)}^2}}}{4} = 1$
${\left( {x + 1} \right)^2} + {\left( {y + 2} \right)^2} = 4$
${\left( {x + 1} \right)^2} + {\left( {y + 2} \right)^2} = {2^2}$
From equation (2),
${\left( {x + 1} \right)^2} + {\left( {y + 2} \right)^2} = {2^2}$is in the form of standard equation of circle
It implies that, $k = \dfrac{3}{4}$ is the correct value.

Option ‘A’ is correct

Note: A circle is a closed curve drawn from a fixed point known as the center, with all points on the curve being the same distance from the center point. Also, to solve such questions don’t check for the first option only. As we got the answer in the first option, it doesn't mean we have to put only one value.