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    Find the Centre and radius of the circle with equation
    x^2+y^2-4x-21=0
    i've got to x^2 -4x+y^2-21 but not sure what to do then any help appreciated
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    Try completing the square for x. [Essentially, you want to get the equation in the form  (x-a)^2 + (y-b)^2 = r^2 ]
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    (Original post by username93)
    Find the Centre and radius of the circle with equation
    x^2+y^2-4x-21=0
    i've got to x^2 -4x+y^2-21 but not sure what to do then any help appreciated
    The standard form of the equation of a circle is this:

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

    When you have it in this form, you can read off the centre as (a,b), and the radius as r.

    So you must get your equation into this format, and the mechanism I would suggest for it would be to complete the square.

    I'll give you a bit of help:

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

    Do the same for the y terms, and then rearrange the equation so you have all the constants on the RHS. It should be apparent from there.
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    im not sure how to do it for y
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    i'm really not sure how to do it for y could someone help
 
 
 
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