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I am given the equation of the circle $x^2+y^2−4x+6y=14$, and I am told to find the equation of the diameter which passes through the origin. However, I am unsure as to how to do this.

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    $\begingroup$ Every diameter passes through the center of the circle. $\endgroup$ – Christopher Carl Heckman Apr 30 '16 at 2:35
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$$x^2+y^2-4x+6y=14$$ $$x^2-4x+4-4+y^2+6y+9-9=14$$ $$(x-2)^2+(y+3)^2=27$$

Hence the center of the circle is $(2,-3)$.

You just have to find the equation of a straight line that connects the origin and the center of the circle.

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