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

  • $\begingroup$ @Frpzzd Yes ofcourse we can do that, but then the Question losts its charm and we do nothing exepth solve the eqn of those circles and then find the eqn of the lines passing through them and the origin and then combine those.......but I wanted to use the theory I wrote in My Approach $\endgroup$ – ami_ba Jun 27 '18 at 19:03
  • $\begingroup$ If you give two extra variabe points you can make several conics pass through their points of intersection.Five points are needed for a conic including degenerate case of a pair of straight lines. $\endgroup$ – Narasimham Jun 27 '18 at 22:52
  • $\begingroup$ I have posted an answer with someone's hint, please comment on the correctness of the method. $\endgroup$ – ami_ba Jun 28 '18 at 7:49

@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:


enter image description here

  • $\begingroup$ The last pair of lines simplifies to $ \dfrac{-3}{8}(x^2+y^2-10 xy/3 )=0$ which is the same result obtained by me but quite directly. $\endgroup$ – Narasimham Jun 29 '18 at 1:23
  • $\begingroup$ The homogenization technique is new to me, up-voted. $\endgroup$ – Narasimham Jun 29 '18 at 1:35

To find the intersection points, solve $$x^2+y^2-2x-2y-2=x^2+y^2-6x-6x+14\implies x+y=4.$$ Now put this into the first equation. $$(4-y)^2+y^2-2(4-y)-2y-2=0\implies y^2-4y+3=(y-1)(y-3)=0$$ so $y=1,3$ giving $x=3,1$ respectively.

Finally, find the equation of the lines as you know $(x_1,y_1)=(0,0)$ and $(x_2,y_2)=(1,3)$ and $(3,1)$

  • $\begingroup$ This is same as what Frpzzd told $\endgroup$ – ami_ba Jun 27 '18 at 19:15

Your approach can’t work because every nontrivial linear combination of two circles is either another circle or a straight line (which you can think of as an infinite-radius circle). As well, the single-parameter family of circles that you have written down doesn’t even include all of the circles that pass through those points—it omits the second of your two circles! A pair of intersecting straight lines, on the other hand, is a degenerate hyperbola, so you have to change the nature of the resulting conic somehow.

  • $\begingroup$ I have used your hint and have posted an answer, please comment on the correctness of the method. $\endgroup$ – ami_ba Jun 28 '18 at 7:39
  • $\begingroup$ Addressed to @amd? $\endgroup$ – Narasimham Jun 28 '18 at 9:31
  • $\begingroup$ Yes, I used, from @amd 's answer, the fact "you have to change the nature of the resulting conic somehow." $\endgroup$ – ami_ba Jun 28 '18 at 9:50

Given circles are

$$x^2+y^2-2x-2y-2=0,\tag1 $$

$$ x^2+y^2-6x-6x+14= 0 \tag2$$

Subtracting 2) from 1) or what is the same as putting $\lambda=-1$ gives equation of the radical axis.

$$ x+y=4 \tag3 $$

which is the line from which equal tangents can be drawn to the circles.

But this is of no use because the radical axis does not pass through the origin and cannot be made to do so.

Points of intersection $(P,Q)$ are found by intersection of $(1) - (3), (2) -(3) $ as

$$ P(x_1,y_1)= (1,3) ; Q(x_2,y_2)=(3,1); \tag4 $$

Pair of straight lines of individual form through origin have to be multiplied to get their paired equation:

$$ y-m_p x=0; y-m_q x=0 \ \tag5 $$

From 4) the respective slopes are

$$ m_p=3, \quad m_q= \frac13 \tag6 $$

enter image description here

So finally required pair of straight lines equation through origin is expressed as:

$$ (y-3 x)(y-x/3)=0 ; x^2+y^2- \frac{10}{3} xy = 0. \tag7 $$


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