Can Three Circles Intersect at a Common Point?

[karush]

studying with a friend there was the intersection of 3 circles problem which is in common usage
here is my overleaf output
View attachment 9075

I was wondering if this could be solved with a matrix in that it has squares in it

or is there a standard equation for finding the intersection of 3 circles given the centers and radius'
and an assumed intersection

 

[HallsofIvy]

Yes, the three circles have equations
$$(x- 7)^2+ (y- 4)^2= 25$$
$$(x+ 9)^2+ (y+ 4)^2= 169$$ and
$$(x+ 3)^2+ (y- 9)^2= 100$$

Multiplying those squares gives
$$x^2- 14x+ 49+ y^2- 8y+ 16= 25$$
$$x^2+ 18x+ 81+ y^2+ 8y+ 16= 169$$
$$x^2+ 6x+ 9+ y^2- 18y+ 81= 100$$

And subtracting will get rid of the squares!

Subtracting the first equation from the second gives
$$32x+ 32+ 16y= 144$$
Subtracting the first equation from the third gives
[tex]20x- 40- 10y+ 65= 75.

32x+ 16y= 112 so 2x+ y= 7
20x- 10y= 50 so 2x- y= 5.
Adding those 4x= 12 so x= 3 and then y= 1.

That, (3, 1), is the point where all three circles intersect.

We also can look at 2x+ y= 7, so y= 7- 2x and $$(x- 7)^2+ (y- 4)^2= (x- 7)^2+ (3- 2x)^2= x^2- 14x+ 49+ 9- 12x+ 4x^2= 5x^2- 26x+ 58= 25$$. $$5x^2- 26x+ 33= 0$$. That can be factored as [tex](5x- 11)(x- 3)= 0[tex] so x= 3 or x= 11/5. If x= 3 y= 7- 6= 1 and if x= 11/5, y= 7- 22/5= (35- 22)/5= 13/5. (3, 1) and (11/5, 13/5) is another intersection.