How to find the equation of diameter of a circle that passes through the origin?
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.
$\endgroup$3 Answers
$\begingroup$Hint: Complete the squares in $x$ and $y$.
$\endgroup$ 1 $\begingroup$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
$Y=M_{OC}(X-X_{0})+Y_{0}$
$y=\frac{-1}{3}(x-0)+0=\frac{-1}{3}x$
$\endgroup$ $\begingroup$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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