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    How do you find the centre of a circle given the radius and one point on it. It appears that I have forgotten everything .

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    Radius is 10. Don't really know what to do with it now.
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    (Original post by Questioness)
    How do you find the centre of a circle given the radius and one point on it. It appears that I have forgotten everything .

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    Radius is 10. Don't really know what to do with it now.
    The formula for the equation of a circle is given by:

    (x-a)^2 + (y-b)^2=r^2

    a and b being the x and y co ordinates of the centre of a circle respectively. TO find the co ordinates try to draw a nice diagram out. You have the radius and a point it touches meaning that you can use pythagoras's theorem to work out the distance, better still put the values into the equation (from he point) and it will give you the centre.
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    (Original post by Maz A)
    The formula for the equation of a circle is given by:

    (x-a)^2 + (y-b)^2=r^2

    a and b being the x and y co ordinates of the centre of a circle respectively. TO find the co ordinates try to draw a nice diagram out. You have the radius and a point it touches meaning that you can use pythagoras's theorem to work out the distance, better still put the values into the equation (from he point) and it will give you the centre.
    Isn't distance = radius ?
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    (Original post by Questioness)
    Isn't distance = radius ?
    yes

    well radius²
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    The OP knows the radius is 10; she is asking what to do from there.

    OP, look at your sketch, what can you say about the line segments?
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    (Original post by Zacken)
    The OP knows the radius is 10; she is asking what to do from there.

    OP, look at your sketch, what can you say about the line segments?
    Makes one long line joined by the two circles?
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    (Original post by Questioness)
    Makes one long line joined by the two circles?
    Joined by *what* of the two circles?

    What is the length of that line? Can you maybe form some sort of useful triangle from that line by using the known centre?
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    (Original post by Zacken)
    Joined by *what* of the two circles?

    What is the length of that line? Can you maybe form some sort of useful triangle from that line by using the known centre?
    Yup, don't really see how Pythagoras is gunna find the centre though. Carry on

    Length of the Line=radius so therefore 10.
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    (Original post by Questioness)
    Yup, don't really see how Pythagoras is gunna find the centre though. Carry on

    Length of the Line=radius so therefore 10.
    Are you the length of the line is 10?

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    (Original post by Zacken)
    Are you the length of the line is 10?

    Ohh, I see. Initially, I drew it like this, why is it wrong?
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    Pictured it at a bad angle sorry. You'll need to tilt it
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    (Original post by Questioness)
    Ohh, I see. Initially, I drew it like this, why is it wrong?
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    That's fine, it's just that the triangle isn't of much use - whereas, in this:



    You can then easily find the angle theta since you know the length of the green line (4-1) and the length of the horizontal black line to the green line (6-2).

    Then \sin \theta = \frac{3}{5} and \cos \theta = \frac{4}{5}.

    Then, you also know that \sin \theta = \frac{3}{5} = \frac{a}{15} So a = 9 but that's the length of the vertical black line, so you gotta add something to it, blah blah, can you see the method and do it for yourself?
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    (Original post by Zacken)
    That's fine, it's just that the triangle isn't of much use - whereas, in this:



    You can then easily find the angle theta since you know the length of the green line (4-1) and the length of the horizontal black line to the green line (6-2).

    Then \sin \theta = \frac{3}{5} and \cos \theta = \frac{4}{5}.

    Then, you also know that \sin \theta = \frac{3}{5} = \frac{a}{15} So a = 9 but that's the length of the vertical black line, so you gotta add something to it, blah blah, can you see the method and do it for yourself?
    I got it. Thanks
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    (Original post by Questioness)
    I got it. Thanks
    No problem - can't believe it took me such a long time to spot the method. Glad you got it.
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    Yea, I'm stuck on another question .

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    Tried using this method
    https://m.youtube.com/watch?v=VZFeyS76euI
    But non the the midpoints for the y axis match the answer.
    Attachment 528969528971
    Answer is (-2,2)
    Attached Images
     
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    (Original post by Questioness)
    Yea, I'm stuck on another question .

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    Tried using this method
    https://m.youtube.com/watch?v=VZFeyS76euI
    But non the the midpoints for the y axis match the answer.
    Attachment 528969528971

    Answer is (-2,2)
    You've found the midpoints - now find the equation of two perpendicular bisectors (one for AB and one for AC) and then equate them to each other to find where they intersect (which is the centre).

    Which is precisely what the video says to do, are you sure you've watched the video?
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    (Original post by Zacken)
    You've found the midpoints - now find the equation of two perpendicular bisectors (one for AB and one for AC) and then equate them to each other to find where they intersect (which is the centre).

    Which is precisely what the video says to do, are you sure you've watched the video?
    Yeah, I have issues.
    Thanks again XD
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    (Original post by Questioness)
    Yeah, I have issues.
    Thanks again XD
    No worries - let me know if you manage it!
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    (Original post by Zacken)
    No worries - let me know if you manage it!

    :nutcase: I got it!! Thank you soooo much.

    Also, could you pick any of the lines eg it could be BC and AB?
    Or does one of it have to be the longest line (AC)

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    (Original post by Questioness)
    :nutcase: I got it!! Thank you soooo much.
    Well done!!
 
 
 
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