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.
$$x^2+y^2-4x+6y=14$$ $$x^2-4x+4-4+y^2+6y+9-9=14$$ $$(x-2)^2+(y+3)^2=27$$
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.