Question

The equation of that diameter of the circle $x^{2}+y^{2}-6x+2y-8=0$ which passes through the origin is

• A. $6x - y = 0$
• B. $3x + 2y = 0$
• C. $x + 3y = 0$
• D. $3x - y = 0$

Answer 1

Answer: (C)

Given equation of a circle is
$x^{2}+y^{2}-6 x+2 y-8=0$
$\therefore $ Centre $=(3,-1)$
Now, the diameter also passes through the origin $O(0,0)$. So, the required equation of the diameter is
$y-0 =\frac{-1-0}{3-0}(x-0)$
$\Rightarrow y=-\frac{1}{3}(x-0)$
$\Rightarrow 3 y=-x$
$\Rightarrow x+3 y=0$