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Which is the correct equation of a circle? Help please :) watch

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    So I'm very rusty on Maths, and I'm just wondering if anyone knew the answer to this?

    *PIC BELOW*

    any help is appreciated.
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    thanks again!
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    Complete the square for x and y to get the equations in the form  (x-a)^2+(y-b)^2=r^2 which recall is the equation for a circle with radius r and centre (a,b).
    Remember that r2 cannot be negative.
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    (Original post by zezno)
    thanks again!
    A circle of centre (a,b) and radius r has the equation
    (x-a)^2 + (y-b)^2 = r^2

    You can eliminate several wrong answers on the basis that r^2 must be positive, it has to x^2 plus y^2 and you won't have an xy term.

    You need to complete the square for x and y terms in c and d to get them into the above format to find the right answer.

    Edit:beaten to it

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    (Original post by gdunne42)
    A circle of centre (a,b) and radius r has the equation
    (x-a)^2 + (y-b)^2 = r^2

    You can eliminate several wrong answers on the basis that r^2 must be positive, it has to x^2 plus y^2 and you won't have an xy term.

    You need to complete the square on c and d to get them into the above format to find the right answer.

    Edit:beaten to it

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    you're beautiful
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    Oh damn, I just realised someone said what I said before me
    I guess I should refresh more often
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    (Original post by zezno)
    you're beautiful
    You're welcome....
    B_9710 was faster


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    (Original post by gdunne42)
    You're welcome....
    B_9710 was faster


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    B_9710

    any idea for this one? I managed to do Q2 and Q3 but a bit confused on this one. Thanks
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    (Original post by zezno)
    B_9710

    any idea for this one? I managed to do Q2 and Q3 but a bit confused on this one. Thanks
    For the first part sub x=2 into the equation and then solve for y.
    Then you have the coordinates of these points. You can find the gradient of the line between the centre of the circle bs the 2 points. Then remember that the tangent to a circle is perpendicular to the radius.
 
 
 
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