Question

One of the diameter of the circle $x^2+y^2-12x+4y+6=0$ is given by

• A. $x+y=0$
• B. $x+3y=0$
• C. $x=y$
• D. $3x+2y=0$

Answer 1

The correct option is B

$x+3y=0$

Explanation for the correct option.

Finding one of the diameters of the circle

Given, the circle is $x^2+y^2-12x+4y+6=0$

$x^2+y^2-12x+4y+6=0$ whose center is $(6,-2).$
We know that center of the circle lies on the diameter of the circle.

Substituting $x=6andy=-2,$ in $x+3y=0$ we get;

$$\begin{gathered} 6+3\times (-2)=0 \\ \Rightarrow 0=0 \end{gathered}$$
Therefore, the point$(6,-2)$ satisfies the equation $x+3y=0$

Hence, the equation of the diameter is $x+3y=0$.

Explanation for the incorrect option:

Substituting $x=6andy=-2,$ in $x+y=0$we get;

$$\begin{gathered} 6+(-2)=0 \\ \Rightarrow 4=0 \end{gathered}$$

Similarly, Substituting $x=6andy=-2,$ in $x=y$we get;

$6\ne -2$

Substituting $x=6andy=-2,$ in $3x+2y=0$we get;

$3.6+2.(-2)\ne 0$

Hence, option (B) is correct.