The origin is a centre of a general conic of second degree iff the coefficients of linear terms vanish.
$ (\Rightarrow)$ part: Let $$ Q(x,y)\equiv ax^{2}+2h xy+ by^{2} + 2gx+2fy+c=0$$ books solution:
let $(x_{1},y_{1})$ be a point on the curve, then $$Q(x_{1},y_{1})=0 \ldots (1)$$ also the point$(-x_{1},-y_{1})$ also lies on the curve , so $$Q(-x_{1},-y_{1})=0 \ldots(2)$$ , subtracting equation $(2) $ from $(1)$ , we get
$$ gx_{1}+fy_{1}=0$$
Now, unless $d,e$ both vanish, $ gx_{1}+fy_{1}=0$ can't be true, 'cause then $x_{1}=0$ and $y_{1}=0$, and it becomes the centre (by hypothesis) , and by definition centre of a curve is a point which bisects every chord passing through it, but with $(x_{1},y_{1})=(0,0)$ this is not possible as centre in this case cannot lie on the curve to have equation $(1)$, so we get sort of contradiction.
THis last part of the proof is what I understood from the book, not what was exactly given in the book.
Now my question is: Why did the author assume initially that $(x_{1},y_{1})$ also lie on the curve? is it because of what was given later in the proof that center bisects every chord passing through it, and $y=x$ is the chord passing through the origin(center) here, so chord must have two common points with the curve, hence the assumption.
Am i right?
Please help me in understanding this part of the proof clearly. It is bugging me to not understand this"easy looking" proof ,
Thanks in advance.
But then it all follows from what you wrote: the center of a quadratic is the origin iff every cord of the quadratic through the origin is bisected by it, which means that if we draw a cord passing through the origin between two points on the curve $\;(x_1,y_1)\;,\;\;(x_2,y_2)\;$ , then $\;(0,0)\;$ is the middle point of the cord iff
$$(0,0)=\left(\frac{x_1+x_2}2\;,\;\;\frac{y_1+y_2}2\right)$$
according to the well know form of the middle point of a line segment, and the above forces
$$x_2=-x_1\;,\;\;y_2=-y_1\;$$
and thus in the proof you mention, assuming the origin is the conic's centre, we certainly have that $\;(x_1,y_1)\;$ on the curve $\;\implies\;(-x_1,-y_1)\;$ also on the curve.