MHB Can Three Circles Intersect at a Common Point?

AI Thread Summary
The discussion focuses on solving the problem of finding the intersection point of three circles defined by their equations. The circles' equations are provided, and a method involving subtraction of these equations is used to eliminate squares and simplify the problem. Through this process, the intersection point (3, 1) is determined, along with an additional intersection point (11/5, 13/5). The conversation highlights the mathematical techniques used, including matrix manipulation and factoring quadratic equations. Overall, the thread demonstrates a systematic approach to solving the intersection of multiple circles.
karush
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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
 
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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
20x- 40- 10y+ 65= 75.<br /> <br /> 32x+ 16y= 112 so 2x+ y= 7<br /> 20x- 10y= 50 so 2x- y= 5.<br /> Adding those 4x= 12 so x= 3 and then y= 1. <br /> <br /> That, (3, 1), is the point where all three circles intersect.<br /> <br /> 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 (5x- 11)(x- 3)= 0so 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.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
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