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    aztect's Avatar
    aztect Posts: 8, Reputation: 1
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    #1

    Mar 30, 2014, 12:44 AM
    Centre and radius of a circle using equation
    Hello I've been looking at this question for awhile but i still don't understand it. To find the centre and radius of the circle given a equation such as
    (x-2)^2+(y-4)^2=8

    What I've tried is
    x^2-4+y^2-16=8
    x^2+y^2=8+16+4
    x^2+y^2 = 28

    So if i wanted the center ill make either x or y =0?
    which gives me the coordinates to the center being; 5.3,5.3? I'm not sure if this is correct just following what I think i know.
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #2

    Mar 30, 2014, 05:55 AM
    When you have an equation of a circle in the form

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

    the center is at (a,b) and the radius is r. So for your equation you have a=2 and b=4, so the center is at (2,4), and radius = sqrt(8).

    Also,when you expand (x-2)^2 you should get (x-2)(x-2) = x^2-4x+4.
    aztect's Avatar
    aztect Posts: 8, Reputation: 1
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    #3

    Apr 1, 2014, 04:39 AM
    Quote Originally Posted by ebaines View Post
    When you have an equation of a circle in the form

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

    the center is at (a,b) and the radius is r. So for your equation you have a=2 and b=4, so the center is at (2,4), and radius = sqrt(8).

    Also,when you expand (x-2)^2 you should get (x-2)(x-2) = x^2-4x+4.
    just to confirm the coordinates for the center of the following my equation is negative, so does that mean it should be (-2,-4)?
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    ebaines Posts: 12,131, Reputation: 1307
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    #4

    Apr 1, 2014, 05:11 AM
    Quote Originally Posted by aztect View Post
    just to confirm the coordinates for the center of the following my equation is negative, so does that mean it should be (-2,-4)?
    No. A circle with center at (-2,4) and radius would have the equation .
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