Homogenising Technique for Lines Joining the Origin to Curve Intersections

A recurring type of problem asks: given a curve $S$ (typically a conic such as a circle, ellipse, or parabola) and a line $L$ that cuts it at two points $P$ and $Q$, find the combined equation of the two lines $OP$ and $OQ$ joining the origin to these intersection points — without first solving for $P$ and $Q$ individually. The homogenising technique does exactly this in one algebraic step.

Let the curve be $S: ax^2+2hxy+by^2+2gx+2fy+c=0$ and the line be $L: lx+my+n=0$ with $n\neq0$, so that $L$ can be rewritten as

Since this quantity equals $1$ at every point of $L$ — and in particular at $P$ and $Q$, which lie on $L$ — we may insert it, raised to whatever power is needed, into $S$ to make every term of $S$ homogeneous of degree 2, without changing the equation's truth at $P$ or $Q$. Concretely, multiply each degree-1 term of $S$ by $\dfrac{lx+my}{-n}$ (raising a degree-1 term to degree 2) and multiply the constant term $c$ by $\left(\dfrac{lx+my}{-n}\right)^2$ (raising a degree-0 term to degree 2):

The resulting equation is homogeneous of degree 2 in $x,y$, and it is satisfied by $P$ and $Q$ (since it agrees with $S=0$ there) — and being homogeneous, it is therefore satisfied by every point on the lines $OP$ and $OQ$, not merely at $P$ and $Q$ themselves. So this homogenised equation is precisely the combined equation of $OP$ and $OQ$, and every earlier result of the chapter — the angle formula $\tan\theta=\left|2\sqrt{h'^2-a'b'}\,/\,(a'+b')\right|$, the perpendicularity test $a'+b'=0$, the coincidence (tangency) test $h'^2=a'b'$ — can be applied directly to its coefficients $a',h',b'$.

As a worked example, take the circle $x^2+y^2=9$ and the line $2x+y=3$, i.e. $\dfrac{2x+y}{3}=1$. Homogenising,

i.e. $x(3x+4y)=0$: the lines from the origin to the two intersection points are $x=0$ and $3x+4y=0$. Here $a'=-3,h'=-2,b'=0$, so $\tan\theta=\left|2\sqrt{4-0}/(-3)\right|=\tfrac43$, giving $\theta=\arctan\tfrac43$ between the two lines — obtained entirely from the coefficients, with no need to solve for $P$ and $Q$ explicitly. …