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avatar+934 

Answer: 3

 Mar 16, 2019
 #1
avatar+934 
0

I tried to graph all 3, but how would you know where they intersect, at one point ot two points without a calc?

 Mar 16, 2019
 #2
avatar+111394 
0

Set the circles equal

 

(x - 3)^2 + (y - 4)^2  =  (x + 4)^2 + (y - 3)^2

x^2 -  6x + 9   + y^2  - 8y + 16  =  x^2 + 8x + 16 + y^2 -6y + 9

-6x - 8y =   8x - 6y

-2y = 14x

y = -7x

 

Sub this into (x - 3)^2 + (y - 4)^2 = 25

(x - 3)^2 + ( -7x - 4)^2 = 25

x^2 - 6x + 9 + 49x^2 + 56x + 16 = 25

50x^2 + 50x + 25 = 25

x ( x + 1) = 0

x = 0     or x = -1

So....the intersection of the circles will occur at   (0,0) and (-1, 7)

 

Find the intersection of    (x - 3)^2 + ( y - 4)^2 = 25   and   y = - 6x

 

(x - 3)^2 + ( -6x - 4) = 25

x^2 - 6x + 9 + 36x^2 + 48x + 16 = 25

37x^2 + 42x  = 0

x (37x + 42) = 0 

x = 0   or x = -42/37

Putting these into y = -6x

(0, 0)  and (-42/37, 252/37)

 

And find the intersection of   (x +4)^2 + ( y - 3)^2 = 25   and   y = - 6x

(x + 4)^2 + (-6x - 3)^2 = 25

x^2 + 8x + 16 + 36x^2 + 36x = 9 = 25

37x^2 + 44x = 0

x (37x + 44) = 0

x = 0   or x = -44/37

Putting these into y = -6x

(0, 0)   and ( -44/37, 264/37)

 

Note that (0,0 )   will satisfy all three

 

But only  (-1, 7) ( -42/37, 252/37)  and (-44/37, 264/37)  will satisfy  two of the three

 

 

cool cool cool

 Mar 16, 2019
 #3
avatar+934 
0

Thanks!!!

dgfgrafgdfge111  Mar 18, 2019

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