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Math Help - Circle Graph?

  1. #1
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    Circle Graph?


    Circles, when graphed on the coordinate plane, have an equation of x2 + y2 = r2 where r is the radius (standard form) when the center of the circle is the origin. When the center of the circle is (h, k) and the radius is of length r, the equation of a circle (standard form) is (x - h)2 + (y - k)2 = r2.

    Example: 1. Problem: Find the center and radius of (x - 2)sq + (y + 3)sq = 16. Then graph the circle.

    Solution: Rewrite the equation in standard form. (x - 2)sq + [y - (-3)]sq = 4sq

    The center is (2, -3) and the radius is 4. The graph is easy to draw, especially if you use a compass. The figure below is the graph of the solution.

    here are my questions:
    ~How did they get -2 and -3 as the center point of the circle?
    ~How come the formula change from x2+y2= r2 to (x-h)2 + (y-k)2 = R2?
    ~Sorry i'm kind of dumb on these things. =) thank you
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  2. #2
    Senior Member yeKciM's Avatar
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    your equation of circle is

     x^2+y^2=r^2

    that will be one with center in (0,0)

    but equation of sifted circle is

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

    so to see where your center is just :

     (x-a) \Rightarrow x-a=0 \Rightarrow  x=a

     (y-b)  \Rightarrow y-b=0 \Rightarrow y=b

    so center is in (a,b)



    both formulas are the same, just in first one ( with center circle) you have that your a=0 and b=0 meaning that it's not shifted ...


    P.S. your center up there is not as you write (-2,-3) it's (2,-3) but that just typo i think
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  3. #3
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    Quote Originally Posted by Gordon View Post

    Circles, when graphed on the coordinate plane, have an equation of x2 + y2 = r2 where r is the radius (standard form) when the center of the circle is the origin. When the center of the circle is (h, k) and the radius is of length r, the equation of a circle (standard form) is (x - h)2 + (y - k)2 = r2.

    Example: 1. Problem: Find the center and radius of (x - 2)sq + (y + 3)sq = 16. Then graph the circle.

    Solution: Rewrite the equation in standard form. (x - 2)sq + [y - (-3)]sq = 4sq

    The center is (2, -3) and the radius is 4. The graph is easy to draw, especially if you use a compass. The figure below is the graph of the solution.

    here are my questions:

    ~How did they get 2 and -3 as the center point of the circle?

    Answer: The equation of a circle is Pythagoras' theorem applied to every point (x,y) on the circle circumference.
    Pick any point on the circumference.
    Using the centre, draw a right-angled triangle.
    The radius of the circle is the hypotenuse of the right-angled triangle.
    The length of the horizontal side is the difference between the x co-ordinates.
    The length of the vertical side is the difference between the y co-ordinates.
    Therefore Pythagoras' theorem gives


    \left(x-x_c\right)^2+\left(y-y_c\right)^2=r^2

    \left(x_c,y_c\right) is the circle centre.

    ~How come the formula change from x2+y2= r2 to (x-h)2 + (y-k)2 = R2?

    Answer: If the centre is (0,0), then Pythagoras' theorem gives

    (x-0)^2+(y-0)^2=x^2+y^2=r^2

    ~Sorry i'm kind of dumb on these things. =) thank you
    .
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  4. #4
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    @Ye, your the best bro.

    @Archie, almost got me confuse but i still got it.

    Thanks both

    ~Gordon
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