Find these equation of the pair of straight lines through the origin which please through the intersection of these curves: $x^2+y^2-2x-2y-2=0$ and $x^2+y^2-6x-6y+14=0$
My approach:
The equation of the family of curves passing through the intersection of the given two curves will be $x^2+y^2-2x-2y-2+\lambda (x^2+y^2-6x-6y+14)=0$
Now, for this curve to be a pair of straight lines passing through the origin, should be a homogenous second degree equation.
So we have to choose a suitable $\lambda$, so that there coefficient of $x$ and $y$ and also the constant term is $0$. But I am unable to do that. Please help
Best Answer
@amd's answer gave me the hint.
First, if we take $\lambda$ to be $-1$ then we get a non homogeneous equation of first degree($x+y=4$) which would surely represent a straight line passing through the points of intersection of the two curves. Now if we somehow homogenize the equation of the second circle withe the help of the equation of the line, i.e $x+y=4$, to a second degree equation, then this equation would surely represent a pair of straight lines through the origin, passing through the points of intersection of the two curves.
I did this by writing $1=\frac{x+y}{4}$ and so multiplying this thing in the RHS with all the terms of first degree in the equation of the second curve and the square of this to the term of zeroth degree, i.e the constant $+14$.
This gives the equation of the pair of straight lines:
$$x^2+y^2-6x\left(\frac{x+y}{4}\right)-6y\left(\frac{x+y}{4}\right)+14\left(\frac{x+y}{4}\right)^2=0$$