Question

The equation of a diameter of circle \[{x^2} + {y^2} - 6x + 2y = 0\] passing through origin is
A. \[x + 3y = 0\]
B. \[x - 3y = 0\]
C. \[3x + y = 0\]
D. \[3x - y = 0\]

Answer 1

Hint: First, we know that the general equation of the circle is \[{x^2} + {y^2} + 2ax + 2by + c = 0\], where \[\left( { - a, - b} \right)\] is the center of the circle. Use this equation, and then use the given conditions to check which of the equations satisfies the obtained point, that is, the center of the circle.

Complete step by step answer
Given the equation of the circle is \[{x^2} + {y^2} - 6x + 2y = 0\].

We know that the general equation of the circle \[{x^2} + {y^2} + 2ax + 2by + c = 0\], where \[\left( { - a, - b} \right)\] is the center of the circle.

Comparing the above equation and the general equation of the circle to find the center of the circle, we get

\[\left( { - a, - b} \right) = \left( {3, - 1} \right)\]

First, we will substitute the value of this center \[\left( {3, - 1} \right)\] in the equation \[x + 3y = 0\]

\[
   \Rightarrow 3 + 3\left( { - 1} \right) = 0 \\
   \Rightarrow 3 - 3 = 0 \\
   \Rightarrow 0 = 0 \\
 \]

Since \[{\text{L.H.S. = R.H.S.}}\], the center of the circle satisfies this equation.

We know that the center of the given circle satisfied the equation \[x + 3y = 0\], but we will also check for the other equation.

So, we will now substitute the value of this center \[\left( {3, - 1} \right)\] in the equation \[x - 3y = 0\].

\[
   \Rightarrow 3 - 3\left( { - 1} \right) = 0 \\
   \Rightarrow 3 + 3 = 0 \\
   \Rightarrow 9 \ne 0 \\
 \]

Since \[{\text{L.H.S. }} \ne {\text{ R.H.S.}}\], the center of the circle does not satisfy this equation.

We will now substitute the value of this center \[\left( {3, - 1} \right)\] in the equation \[3x + y = 0\].

\[
   \Rightarrow 3\left( 3 \right) + \left( { - 1} \right) = 0 \\
   \Rightarrow 9 - 1 = 0 \\
   \Rightarrow 8 \ne 0 \\
 \]

Since \[{\text{L.H.S. }} \ne {\text{ R.H.S.}}\], the center of the circle also does not satisfy this equation.

We will now substitute the value of this center \[\left( {3, - 1} \right)\] in the equation \[3x - y = 0\].

\[
   \Rightarrow 3\left( 3 \right) - \left( { - 1} \right) = 0 \\
   \Rightarrow 9 + 1 = 0 \\
   \Rightarrow 10 \ne 0 \\
 \]

Since \[{\text{L.H.S. }} \ne {\text{ R.H.S.}}\], the center of the circle also does not satisfy this equation.

Therefore, the option A is correct.

Note: In solving these types of questions, we have to compare the given equation with general equation of the circle \[{x^2} + {y^2} + 2ax + 2by + c = 0\], where \[\left( { - a, - b} \right)\] is the center of the circle to find the center of the circle easily. After that, the question will become really simple to solve by just substituting the value of the center in the given options.