A non-intersecting family of circles

  • #1
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Member warned that homework must be posted in one of the Homework sections
Mentor note: Moved from a technical math section.
What is the proof that the family of circles out of two non-intersecting circles, no two circles in that family intersect.

Say S1 = x^2 + y^2 - 8x + 7 = 0 (i.e center at (4,0) and radius = 3 )
S2 = x^2 + y^2 - 24x + 135 = 0 ( i.e center at (12,0) and radius = 3 )
Family of circles of the two above circles is S1 + k S2 = 0.
i.e x^2 + y^2 + [2* (-4 -12k)/(1+k)] x + [ (7 + 135k ) / (1+k) ] = 0 , k ∈ R

Thanks.
 
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Answers and Replies

  • #2
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Did you try to find x,y that satisfy the conditions for both circles? Or did you calculate the centers and radii of the circles as function of k?
 
  • #3
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Did you try to find x,y that satisfy the conditions for both circles? Or did you calculate the centers and radii of the circles as function of k?
For circles along x-axis.

S1 = x^2 + y^2 + 2g1x + c1 = 0
S2 = x^2 + y^2 + 2g2x + c2 = 0

Family of circle of the above two circle. Center and radii as function of k.
center = ( - ((g1+kg2)/(1+k)) , 0 ) and radius = √ ( [(g1+kg2)/(1+k)]^2 - [(c1+kc2)/(1+k)] )

From my example, it is.

Center = ( -[{-4(1+3k)}/(1+k)] , 0) and radius = √ ( [{-4(1+3k)}/(1+k)]^2 - [(7+135k)/(1+k)] )

How do I proove from here?
 
  • #4
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How do I proove from here?
It is your homework problem. How would you check if these circles have a common point?

One possible approach: It helps that all their centers are aligned. What do you know about the intersections with the x axis for the circles? How must their arrangement be like if the circles intersect?
 

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