Find the equation a pair of straight lines through origin which passes through the intersection of curves whose equation are $$L1: x^2+y^2-2x-2y-2=0$$ and $$L2: x^2+y^2-6x-6y+14=0$$.

I know that the curves have to be homogenised, although I don't know how to homogenise a curve with another. All I know is how to homogenise a curve with line, but this is completely different. Please help.


Subtract the two equations $L1-L2$ will give

$x^2+y^2-2x-2y-2-(x^2+y^2-6x-6y+14)=0\rightarrow 4 x + 4 y=16 \rightarrow y=4-x$

Substitute in $L1$

$x^2+(4-x)^2-2x-2(4-x)-2=0\rightarrow 2 x^2-8 x+6=0 $

and finally the two points of intersection of the two circles $A(1;\;3),\;B(3;\;1)$

Then is easy to write the equation of the lines that pass through the origin and $A$ and $B$

Hope this helps

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