0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.