Question

Equation of the diameter of the circle $x^2+y^2-12x+4y+3=0$ 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 $x+3y=0$

Equation of the circle is

$x^2+y^2-12x+4y+3=0⟶\left(1\right)$

The standard equation of a circle is

$x^2+y^2+2gx+2fy+c=0$

where the center is $(-9,-f)$ and radius is $\sqrt{9^2+f^2-c}$

$\therefore$ From $\left(1\right)$

the center of the circle is $(6,-2)$ and radius will be $\sqrt{36+4-3}=\sqrt{37}$ units

Now the diameter will contain the center. Hence the equation of the diameter should be satisfied by the center $(6,-2)$

Check the options :

A. $x+y=0$

Putting $x=6$ $y=-2$ we get

$6-2=0\Rightarrow 4=0$, which is not true

B. $x+3y=0$

Putting $x=6$ $y=-2$ we get

$6+3x-2=0$

$\Rightarrow 6-6=0\Rightarrow 0=0$

Hence the correct answer will be ${}^{\prime }{B}^{\prime }$

i.e. the equation of diameter will be $x+3y=0$