The equation of a diameter of the circle x^2 | vedclass

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Question

(A) The given equation of the circle is $x^2 + y^2 - 6x + 2y = 0$.Comparing this with the general equation $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $2g

Vedclass · Accepted Answer

$x + 3y = 0$

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(A) The given equation of the circle is $x^2 + y^2 - 6x + 2y = 0$.Comparing this with the general equation $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $2g = -6 \implies g = -3$ and $2f = 2 \implies f = 1$.The center of the circle is $(-g, -f) = (3, -1)$.$A$ diameter of the circle always passes through its center.We need the equation of a line passing through the origin $(0, 0)$ and the center $(3, -1)$.The slope $m$ of the line is $\frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - 0}{3 - 0} = -\frac{1}{3}$.The equation of the line is $y - 0 = -\frac{1}{3}(x - 0)$,which simplifies to $3y = -x$ or $x + 3y = 0$.

The equation of a diameter of the circle $x^2 + y^2 - 6x + 2y = 0$ passing through the origin is:

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