Where Do Chords of Circles Through A(3,7) and B(6,5) Intersect?

[Saitama]

Homework Statement

Consider a family of circles passing through two fixed points A(3,7) and B(6,5). Show that the chords in which the circle $x^2+y^2-4x-6y-3=0$ cuts the members of the family are concurrent at a point. Find the coordinates of this point.

Homework Equations

The Attempt at a Solution

The family of circles passing through the two given points is given by:
$$(x-1)(x-6)+(y-7)(y-5)+\lambda \left|
\begin{array}{c c c}
x & y & 1\\
3 & 7 & 1\\
6 & 5 & 1\\
\end{array}
\right|=0$$
I am completely clueless about the next step.

Any help is appreciated. Thanks!

 

[HallsofIvy]

An obvious first step would be to multiply out that last equation to the form x^2+ y^2+ ax+ by + c= 0, in terms of \lambda, then solve that equation together with x^2+ y^2- 4x- 6y- 3= 0.

 

[Saitama]

An obvious first step would be to multiply out that last equation to the form x^2+ y^2+ ax+ by + c= 0, in terms of \lambda, then solve that equation together with x^2+ y^2- 4x- 6y- 3= 0.

I don't think that would be a good idea (I did think of this initially but looking at the equations formed, I left the approach).

Rewriting the last equation, I get:
$$x^2+y^2+x(2\lambda-9)+y(3\lambda-12)+53-27\lambda=0$$

Looks like it won't be easy to solve this way. Any other ideas?

 

[haruspex]

[STRIKE]Maybe I made an error, but it doesn't seem to be true. Are you sure you've quoted all the constants correctly?[/STRIKE]
Anyway, I suggest simplifying the equations by a few shifts in co-ordinates. You can universally replace x by x+2 and y by y+3 to centre the fixed circle on the origin. Next, you can similarly apply any affine shift to lambda without affecting anything (adding 2 looks good).
For the intersection between the fixed circle and any of the family, you can arrange to eliminate the quadratics. What do you think the resulting linear equation represents?

Edit: Error found.

 

[Saitama]

Hi haruspex! :)

[STRIKE]Maybe I made an error, but it doesn't seem to be true. Are you sure you've quoted all the constants correctly?[/STRIKE]
Anyway, I suggest simplifying the equations by a few shifts in co-ordinates. You can universally replace x by x+2 and y by y+3 to centre the fixed circle on the origin. Next, you can similarly apply any affine shift to lambda without affecting anything (adding 2 looks good).
For the intersection between the fixed circle and any of the family, you can arrange to eliminate the quadratics. What do you think the resulting linear equation represents?

Edit: Error found.

Finding out the common chord didn't hit me, thanks a lot! :)

But I think there is no need of shifting the origin. The common chord for the fixed circle and the family is given by:
$$(2\lambda-5)x+(3\lambda-6)y+56-27\lambda=0$$
This is a family of straight lines concurrent at a point. Substituting $\lambda=2$ and $\lambda=5/2$ gives the desired point.

Thanks a lot haruspex!

 

[haruspex]

Hi haruspex! :)



Finding out the common chord didn't hit me, thanks a lot! :)

But I think there is no need of shifting the origin. The common chord for the fixed circle and the family is given by:
$$(2\lambda-5)x+(3\lambda-6)y+56-27\lambda=0$$
This is a family of straight lines concurrent at a point. Substituting $\lambda=2$ and $\lambda=5/2$ gives the desired point.

Thanks a lot haruspex!

You're welcome. It wasn't obvious to me there was a common point without making the substitutions. How did you show it?

 

[Saitama]

You're welcome. It wasn't obvious to me there was a common point without making the substitutions. How did you show it?

I rewrite the linear equation as:
$$-5x-6y+56+\lambda(2x+3y-27)=0$$
I don't see what else it can represent except the family of lines concurrent at a point.

 

[haruspex]

I rewrite the linear equation as:
$$-5x-6y+56+\lambda(2x+3y-27)=0$$
I don't see what else it can represent except the family of lines concurrent at a point.

OK, but by what reasoning?
E.g. $-4x-6y+56+\lambda(2x+3y-27)=0$ does not represent a family of lines concurrent at a point.

 

[Saitama]

OK, but by what reasoning?
E.g. $-4x-6y+56+\lambda(2x+3y-27)=0$ does not represent a family of lines concurrent at a point.

I am not sure how to give a reason. This is what I think:

Let
$L_1: -5x-6y+56=0$
$L_2: 2x+3y-27=0$

The point of intersection of these two lines is $(2,23/3)$. Any line represented by $L_1+\lambda L_2=0$, where $\lambda$ is a parameter, obviously passes through (2,23/3). Is this enough or should I add something more?

 

[haruspex]

I am not sure how to give a reason. This is what I think:

Let
$L_1: -5x-6y+56=0$
$L_2: 2x+3y-27=0$

The point of intersection of these two lines is $(2,23/3)$. Any line represented by $L_1+\lambda L_2=0$, where $\lambda$ is a parameter, obviously passes through (2,23/3). Is this enough or should I add something more?

That works fine. Note that for my example with -4 instead of -5 there is no solution to the pair of equations.

 

[Saitama]

Note that for my example with -4 instead of -5 there is no solution to the pair of equations.

Yes, I did notice that. Thank you very much haruspex! :)

 

[TheScholarAM]

OK, but by what reasoning?
E.g. $-4x-6y+56+\lambda(2x+3y-27)=0$ does not represent a family of lines concurrent at a point.

Oh hey dude you've got it wrong. They are two parallel lines they don't intersect at a point they can never represent a family of lines passing through their intersection point.

 

[TheScholarAM]

Oh hey dude you've got it wrong. They are two parallel lines they don't intersect at a point they can never represent a family of lines passing through their intersection point.

The concept here is that a line passes through the intersection of 2 lines of the point satisfies the equation of the line right. It is always of the form L1+xL2 as every line represented by this equation has the same common root as the intersection point that satisfies it. Hope you get a better picture of it now.

 

[haruspex]

they don't intersect at a point

Isn’t that what I wrote?