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Math Help - Finding equation of circle.

  1. #1
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    Finding equation of circle.

    Ok so it says on the question:

    Write the equation of the following circle:
    with diameter AB and given A(4,3) and B(6,-1)

    please need help really fast
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  2. #2
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    Quote Originally Posted by behindDkeys View Post
    Ok so it says on the question:

    Write the equation of the following circle:
    with diameter AB and given A(4,3) and B(6,-1)

    please need help really fast
    The mid point is the center of the circle

    (x-x_0)^2+(y-y_0)^2=r^2
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    Quote Originally Posted by behindDkeys View Post
    Ok so it says on the question:

    Write the equation of the following circle:
    with diameter AB and given A(4,3) and B(6,-1)

    please need help really fast
    Find the distance between these two points using Pythagoras' Theorem. Halve it and you have the radius ( r).

    Find the midpoint of the segment between these two points. This is your centre (h, k).


    The formula of the circle is

    (x - h)^2 + (y - k)^2 = r^2.
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    Quote Originally Posted by Prove It View Post
    Find the distance between these two points using Pythagoras' Theorem. Halve it and you have the radius ( r).

    Find the midpoint of the segment between these two points. This is your centre (h, k).


    The formula of the circle is

    (x - h)^2 + (y - k)^2 = r^2.

    ok so what i do is i take the points and put it on an A^2+B^2=C^2?


    coz i dont really get pythagoras on this i dont see ABC
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  5. #5
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    Once you have the mid point, you can figure out the distance to one of the points using the distance formula. The distance you will have solved for will be the radius.
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    Quote Originally Posted by behindDkeys View Post
    ok so what i do is i take the points and put it on an A^2+B^2=C^2?


    coz i dont really get pythagoras on this i dont see ABC
    The distance formula

    D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

    is a direct application of Pythagoras' Theorem.
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  7. #7
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    Quote Originally Posted by dwsmith View Post
    Once you have the mid point, you can figure out the distance to one of the points using the distance formula. The distance you will have solved for will be the radius.
    uh can you show me step by step
    like plug in the given then ima solve the other stuff
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    Quote Originally Posted by Prove It View Post
    The distance formula

    D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

    is a direct application of Pythagoras' Theorem.

    oh crap i used that and accidentally exchanged x_2 with x_1 and also for the y

    thats why i keep getting this square root of 20 =.=
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  9. #9
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    You need to use:


    • The Distance Formula

    D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

    which is a direct application of Pythagoras' Theorem.


    • The Midpoint Formula

    (x_m, y_m) = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right).
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  10. #10
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    Quote Originally Posted by behindDkeys View Post
    oh crap i used that and accidentally exchanged x_2 with x_1 and also for the y

    thats why i keep getting this square root of 20 =.=
    \sqrt{20} = 2\sqrt{5} is correct.
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  11. #11
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    Quote Originally Posted by behindDkeys View Post
    oh crap i used that and accidentally exchanged x_2 with x_1 and also for the y

    thats why i keep getting this square root of 20 =.=

    Doesn't matter what order you go in as long as you are consistent.
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  12. #12
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    Quote Originally Posted by Prove It View Post
    \sqrt{20} = 2\sqrt{5} is correct.
    uhm ok so whats next to that
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    Quote Originally Posted by behindDkeys View Post
    Ok so it says on the question:

    Write the equation of the following circle:
    with diameter AB and given A(4,3) and B(6,-1)

    please need help really fast
    If you are familiar with the properties of circle, try this method.

    Let P (x. y) be any point on the circle. Then APB is a right angled triangle. AP is perpendicular to PB.
    If m1 and m2 are the slopes of AP and PB then m1*m2 = -1.

    So \frac{(y - y_1)}{(x - x_1)} \times \frac{(y - y_2)}{(x - x_2)} = -1

    Expand and get the equation of the circle.
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  14. #14
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    Quote Originally Posted by behindDkeys View Post
    uhm ok so whats next to that

    Hello behindDkeys,
    You have already been given the procedures but appear to need more help.
    Try this.
    Plot the two points and draw AB. AB is the diameter of a circle and its midpoint is the center.Draw the slope diagram for AB.Note that the rise is 4 and the run is 2.Circle diameter is 2rad5 and radius is rad5.Using similar triangles the midpoint of AB is 5,1. Now write the circle equation as shown by others.


    bjh
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  15. #15
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    Quote Originally Posted by bjhopper View Post
    Hello behindDkeys,
    You have already been given the procedures but appear to need more help.
    Try this.
    Plot the two points and draw AB. AB is the diameter of a circle and its midpoint is the center.Draw the slope diagram for AB.Note that the rise is 4 and the run is 2.Circle diameter is 2rad5 and radius is rad5.Using similar triangles the midpoint of AB is 5,1. Now write the circle equation as shown by others.


    bjh
    sorry i forgot to post what i got

    here's the equation i got:
    x^2+y^2=\sqrt{5}

    since radius is divided by 2 and the diameter i got was 2\sqrt{5}

    did i get it right?
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