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anonymous

  • one year ago

I need help with this question...

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  1. anonymous
    • one year ago
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    Prove that the two circles shown below are similar.

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  2. anonymous
    • one year ago
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    i keep seeing these and am totally confused aren't all circles similar?

  3. anonymous
    • one year ago
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    |dw:1440510136349:dw| notice that E was moved to C so now we have a dilation of \[\huge \frac{ r_1 }{ r_2 }\] which gives the scale factor.

  4. anonymous
    • one year ago
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    yes, you need to use a formula to move one over another and dilate it, but idk the formula

  5. anonymous
    • one year ago
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    Sorry about using the same color, but essentially just using a transformation this actually proves all circles are similar to each other.

  6. anonymous
    • one year ago
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    so, what formula would i use to translate it?

  7. anonymous
    • one year ago
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    I mean we could say \[\pi = \frac{ C }{ 2r }\] making all circles the same

  8. anonymous
    • one year ago
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    and would the scale factor be 3/4?

  9. anonymous
    • one year ago
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    4/3

  10. anonymous
    • one year ago
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    ohh, alright

  11. anonymous
    • one year ago
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    So the trick here is knowing that translations and dilations are the same hence all circles are the same

  12. anonymous
    • one year ago
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    i found a new formula for translations

  13. anonymous
    • one year ago
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    \[(x - h){^2} + (y - k){^2} = r{^2}\]

  14. anonymous
    • one year ago
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    That's the equation of a circle

  15. anonymous
    • one year ago
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    oh

  16. anonymous
    • one year ago
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    You may use it though

  17. anonymous
    • one year ago
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    ok so how would i put this into words? each step i mean

  18. mathmate
    • one year ago
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    Perhaps try this: given circle E : (x-4)^2+(y-9)^2=3^2 h=-7, k=-8, scale factor = 4/3 so (x-4+7)^2+(y-9+8)=(3*(4/3))^2 gives (x+3)^2+(y-1)^2=4^2 which is the circle C

  19. anonymous
    • one year ago
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    \[(x,y) \implies (x-h,y-k)\] and end up doing a dilation

  20. anonymous
    • one year ago
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    ok thanks you guys, i have more questions i'll post in a couple of minutes, i'd appreciate it if you could help me with those as well :)

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