# Family of Circles problem

1. Nov 29, 2013

### Saitama

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. 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!

2. Nov 29, 2013

### HallsofIvy

Staff Emeritus
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 $x^2+ y^2- 4x- 6y- 3= 0$.

3. Nov 29, 2013

### Saitama

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?

4. Nov 29, 2013

### 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.

Last edited: Nov 29, 2013
5. Nov 30, 2013

### Saitama

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!

6. Nov 30, 2013

### haruspex

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

7. Nov 30, 2013

### Saitama

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.

8. Nov 30, 2013

### haruspex

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.

9. Dec 1, 2013

### Saitama

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?

Last edited: Dec 1, 2013
10. Dec 1, 2013

### haruspex

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

11. Dec 1, 2013

### Saitama

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