Question

The equation of the diameter of the circle \[3\left( {{x^2} + {y^2}} \right) - 2x + 6y - 9 = 0\] which is perpendicular to line $2x + 3y = 12$ is
(A) $3x - 2y = 3$
(B) $3x - 2y + 1 = 0$
(C) $3x - 2y = 9$
(D) none of these

Answer 1

Hint: First we will convert the equation of circle in its standard format in which the coefficient of ${x^2}$ and ${y^2}$ would be one, and then we will find the center of the circle by comparing the equation of the circle with the standard equation. Then we will find the slope of the required line by using the formula ${m_1}{m_2} = - 1$, where ${m_1}$ is the slope of the required line and ${m_2}$is the slope of the line $2x + 3y = 12$.

Complete step-by-step answer:
First, we will draw the figure according to the given question,

First, we will write the equation of circle in its standard form.
In standard form, the coefficient of ${x^2}$ and ${y^2}$ would be one.
So, the given equation is \[3\left( {{x^2} + {y^2}} \right) - 2x + 6y - 9 = 0\]
Now, we will divide the whole equation by $3$.
\[ \Rightarrow {x^2} + {y^2} - \dfrac{2}{3}x + \dfrac{6}{3}y - \dfrac{9}{3} = 0\]
Now, we will compare the above equation with the standard equation of circle \[{x^2} + {y^2} + 2gx + 2fy + c = 0\]in which the center of the circle is $( - g, - f)$.
So, we will divide the coefficient of x and y with $ - 2$to find the center.
Therefore, the center of the circle is $\left( {\dfrac{{ - 2}}{3} \times \dfrac{1}{{ - 2}},\,\,\dfrac{6}{3} \times \dfrac{1}{{ - 2}}} \right)$
$ \Rightarrow \left( {\dfrac{1}{3}, - 1} \right)$
We all know that the diameter passes through the center. So, the required line passes thought the above point.
Now, we will find the slope of the diameter.
It is given that the diameter is perpendicular to the line $2x + 3y = 12$.
Now, we will use the relation ${m_1}{m_2} = - 1$to find the slope of the required line, here ${m_1}$ is the slope of the required line and ${m_2}$is the slope of the line $2x + 3y = 12$.
${m_2} = \dfrac{{ - coefficient\,of\,x}}{{coefficient\,of\,y}} = \dfrac{{ - 2}}{3}$
Now, on substituting the above relation, we get
${m_1}\left( {\dfrac{{ - 2}}{3}} \right) = - 1$
${m_1} = \dfrac{3}{2}$
Now, we will use the formula of equation of line to find it and its formula is $y - {y_1} = {m_1}\left( {x - {x_1}} \right)$, where ${x_1}\,and\,{y_1}$are the points through which the line is passing. In this case it is the center of the circle.
Therefore, ${x_1} = \dfrac{1}{3},\,{y_1} = - 1$
$y - \left( { - 1} \right) = \dfrac{3}{2}\left( {x - \dfrac{1}{3}} \right)$
$ \Rightarrow y + 1 = \dfrac{3}{6}\left( {3x - 1} \right)$
$ \Rightarrow y + 1 = \dfrac{1}{2}\left( {3x - 1} \right)$
$ \Rightarrow 2y + 2 = 3x - 1$
$ \Rightarrow 3x - 2y = 3$
Therefore, the equation of diameter is $3x - 2y = 3$.

So, the correct answer is “Option A”.

Note: In the above question, we can also find the slope of the line $2x + 3y = 12$ by first converting it in the intercept form in which the coefficient of y should be one and the coefficient of x denotes the slope of that line and the constant term denotes the y intercept of that line.