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    the circle has the equation
    x^2 + y^2 + 2X - 4Y - 3 = 0

    I found the co-ords of the centre of the circle (1,-2) but I don't know how to find the radius without a point on the circumference?
    help please.
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    (Original post by hopcyn)
    the circle has the equation
    x^2 + y^2 + 2X - 4Y - 3 = 0

    I found the co-ords of the centre of the circle (1,-2) but I don't know how to find the radius without a point on the circumference?
    help please.
    From Pythagoras' theorem  (x-a)^2 + (y-b)^2= r^2

    (if you think about it, x-a is the horizontal distance from the centre to any variable point on the locus (circle) and y-b is the vertical distance. if you draw the horizontal distance followed by the vertical distance you can form a right angled triangle with hypotenuse = radius)

    Where (a,b) are the coordinates of the centre of the circle and r is the radius of the circle.

    If you can express your equation in this form then you can compare and find the radius.
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    (Original post by NotNotBatman)
    From Pythagoras' theorem  (x-a)^2 + (y-b)^2= r^2

    (if you think about it, x-a is the horizontal distance from the centre to any variable point on the locus (circle) and y-b is the horizontal distance.
    I think you mean veritcal. :-)

    -----------------------------------------------------------------------------------

    For the original poster:
    - Look at your point of centre again.

    - You can also apply the complete the square method on your equation of the circle to obtain an alternative equation of the circle where the centre and the radius can be read off easily.
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    (Original post by simon0)
    I think you mean veritcal. :-)

    -----------------------------------------------------------------------------------

    For the original poster:
    - Look at your point of centre again.

    - You can also apply the complete the square method on your equation of the circle to obtain an alternative equation of the circle where the centre and the radius can be read off easily.
    Yes, I did, thank you.
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    (Original post by NotNotBatman)
    Yes, I did, thank you.
    No problem. :-)
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    x2+ y2+ 2X - 4Y - 3 = 0

    ( x - a )2 + ( y - b )2 = r2

    can be expanded and rearranged as

    x2 -2ax + y2- 2by +a2+ b 2 - r2 = 0

    compare it to

    x2+ 2x + y2- 4Y - 3 = 0

    and you can find suitable values for a & b
 
 
 
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