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LynFran

  • one year ago

Show that (7,2) and (1,-6) are on a circle whose center is (4,-2) and find the length of the radius

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  1. anonymous
    • one year ago
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    Do you know the standard equation for a circle?

  2. LynFran
    • one year ago
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    x^2+y^2=r^2 ??

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

  4. anonymous
    • one year ago
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    Where (h, k) are the center point

  5. LynFran
    • one year ago
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    ok with (h,k) being the center and r is the radius ..what nxt..

  6. anonymous
    • one year ago
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    Pick a point, (7,2): It can be odd sometimes, because the values may change sign

  7. LynFran
    • one year ago
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    (x-4)^2+(y+2)^2=r^2

  8. LynFran
    • one year ago
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    ok so (3)^2+(4)^2=r^2 9+16=r^2 25=r^2 5=r ?

  9. anonymous
    • one year ago
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    So check if the points are in the circle

  10. LynFran
    • one year ago
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    idk how to...??

  11. LynFran
    • one year ago
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    which point??

  12. anonymous
    • one year ago
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    Hmm, I think I did something wrong - (7,2) is in the circle, but not (1, -6)

  13. LynFran
    • one year ago
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    but we have to show that those points are on the circle

  14. anonymous
    • one year ago
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    hold on

  15. LynFran
    • one year ago
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    ok

  16. anonymous
    • one year ago
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    sorry, I can't help you. It's been too long

  17. LynFran
    • one year ago
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    @Nnesha

  18. LynFran
    • one year ago
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    @campbell_st

  19. campbell_st
    • one year ago
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    ok... so the general form is \[(x - h)^2 + (y - k)^2 = r^2\] where (h, k) is the centre and r is the radius.. so if you substitute your centre you get \[(x - 4)^2 + (y + 2)^2 = r^2\] is that ok so far...?

  20. LynFran
    • one year ago
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    yes

  21. campbell_st
    • one year ago
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    and reading the notes you have done the correct calculation so picking the point (7, 2) and substituting that you can find r^2 \[(7 - 4)^2 + (2 + 2)^2 = r^2\] so \[r^2 = 25\] so just take the square root of 25 to get the radius

  22. campbell_st
    • one year ago
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    it doesn't matter which point on the circle you use, as you'll get the same value so that just means the equation is \[(x - 4)^2 + (y + 2)^2 = 25\]

  23. campbell_st
    • one year ago
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    hope it makes sense

  24. campbell_st
    • one year ago
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    f you want to check you can find the distance from the centre to the points using the distance formula and show the distance is 5 units

  25. LynFran
    • one year ago
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    but they say show that the points (7,2) and (1,-6) are on the circle... i think thats what confuses me... idk how to show this...

  26. campbell_st
    • one year ago
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    sure so you know \[(7 - 4)^2 + (2 + 2)^2 = r^2\] using the other point \[(1 - 4)^2 + (-6 + 2)^2 = r^2\] and you'll find \[r^2 = r^2 ~~~or~~~~25 = 25\]

  27. LynFran
    • one year ago
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    so \[\sqrt{(7-1)^2+(2+6)^2}=10\]

  28. campbell_st
    • one year ago
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    so since both points are the same distance from the centre, by definition, the points are on a circle centre (4, -2) and radius 5

  29. LynFran
    • one year ago
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    oh \[\sqrt{(4-1)^2+(-2+6)^2}=5\]

  30. campbell_st
    • one year ago
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    I wouldn't find the distance between the points. I'd use distance from the centre to each point... which is really the same as substituting into the general form to show that the radius is equal. A simple definition of a circle is a set of points equidistant from a fixed point. so showing that the radius is 5 from both points to a fixed point( the centre) then the 2 points are on a circle, centre (4, -2) and radius 5

  31. LynFran
    • one year ago
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    and \[\sqrt{(4-7)^2+(-2-2)^2}=5\]

  32. campbell_st
    • one year ago
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    that works.... so you have shown the points are the same distance from the fixed point

  33. LynFran
    • one year ago
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    cool thanks

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