How to find the equation of diameter of a circle that passes through the origin?

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So this was a question that I was solving that got me stuck. Its as follows: Q. Find equation of diameter of the circle $x^2 + y^2 - 6x + 2y = 0$ which passes through the origin.

Now I have tried the following,

From the equation I found, $g=(-3)$ and $f=1$ and subsequently found the radius as $r=\sqrt{10}$ and hence diameter as $d=2\sqrt{10}$.

But I cant seem to understand how to progress, help would be greatly appreciated.

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Hint: Complete the squares in $x$ and $y$.

0
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You know the centre of the circle and you have the point $(0,0)$ (which lies on the circle). Those are two points in $\mathbb R^2$. Now just calculate the distance between those two points.

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Let centre of the given circle be $C(x_{1},y_{1})$

Hint Diameter passes through the centre of the circle as well as the origin

$C(x_{1},y_{1})$= $\frac{-1}{2}$$(-6,2)$

$\implies$=$(3,-1)$

Origin, $O(x_{0},y_{0})$=$(0,0)$

$M_{OC}$=slope of diameter=$\frac{(0- -1)}{(0-3)}$

$\implies$$M_{OC}$=$\frac{-1}{3}$

Equation of diameter

$Y=M_{OC}(X-X_{0})+Y_{0}$

$y=\frac{-1}{3}(x-0)+0=\frac{-1}{3}x$