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    x^2+y^2-6-k=0 has radius 4. Find centre of the circle and value of k
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    (Original post by daryl.c)
    x^2+y^2-6-k=0 has radius 4. Find centre of the circle and value of k
    What have you tried?
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    its given me k=-16 and c (0,3) but my friend got something else
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    (Original post by daryl.c)
    its given me k=-16 and c (0,3) but my friend got something else
    This is incorrect, however, did you post the correct equation in your original post? Did you mean to say x^2+y^2-6y-k=0?
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    yeah its just dont have -6y its just -6
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    (Original post by daryl.c)
    yeah its just dont have -6y its just -6
    So then your centre is off. Clearly is must be (0,0).

    Then the equation can be rearranged to say x^2+y^2=6+k and comparison to the general equation of the circle (x-a)^2+(y-b)^2=r^2 should be able to give you the appropriate equation concerning the radius and k.
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    thx mate you got the same as my friend so it must be right c (0,0) with k=10?
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    (Original post by daryl.c)
    thx mate you got the same as my friend so it must be right c (0,0) with k=10?
    That's right.
 
 
 
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