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How to prove point lies in/outside circle? (C2)

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    Hi -

    Please can someone tell me how I can determine whether a point lies inside or outside a circle (with the equation of the circle?)

    Thanks
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    Put the x coordinate of the point into the equation and show that it is not equal to the y value of the point you have been given?
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    (Original post by John Locke)
    Put the x coordinate of the point into the equation and show that it is not equal to the y value of the point you have been given?
    I don't understand lol...
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    Eq of circle ---- x^2+y^2=r^2.

    If for a point P=(a,b),

    a^2+b^2 \leq r^2

    then P lies in (or on the boundary of) the circle.
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    i think its something to do with the radius.
    so find the centre, and if the point to the centre is greater in length (of the radius) then the point lays outside of the circle
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    (Original post by Darkest Knight)
    I don't understand lol...
    Woops, my bad. Misread the question.
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    (Original post by 0-))
    Eq of circle ---- x^2+y^2=r^2.

    If for a point P=(x_1,y_1),

    x_1^2+y_1^2 \leq r^2

    then P lies in (or on the boundary of) the circle.
    Thanks, i don't undersatnd the x (2/1) + y (2/1) if you could explain that for me please?

    Bit confused too, as seem to be different methods?
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    (Original post by 0-))
    Eq of circle ---- x^2+y^2=r^2.

    If for a point P=(x_1,y_1),

    x_1^2+y_1^2 \leq r^2

    then P lies in (or on the boundary of) the circle.
    or for a more general circle, not necessarily centred at the origin

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

    we can just say, given some random point  P(x_1,y_1) if PC < r then it is inside, and if PC > r, then it is outside. where C is the centre of the circle (a,b)

    OP: think about how a circle works: all the points on the circumference are a fixed distance from the centre C. if we are given some random point, and asked to state whether it is inside the circle, outside the circle, or on the circumference, then all we need to do is find the distance of that point from the centre of the circle....
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    that 2 1 is just the subscript of using the point x1 so its the same as the first equation of the circle just using the co-ordinates out of the point p
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    (Original post by Darkest Knight)
    Thanks, i don't undersatnd the x (2/1) + y (2/1) if you could explain that for me please?

    Bit confused too, as seem to be different methods?
    I edited my post to make it less confusing. Can you understand it now?
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    (Original post by GHOSH-5)
    or for a more general circle, not necessarily centred at the origin...
    Oops. I shouldn't have assumed the circle was centred at the origin.
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    (Original post by GHOSH-5)
    or for a more general circle, not necessarily centred at the origin

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

    we can just say, given some random point  P(x_1,y_1) if PC < r then it is inside, and if PC > r, then it is outside. where C is the centre of the circle (a,b)

    OP: think about how a circle works: all the points on the circumference are a fixed distance from the centre C. if we are given some random point, and asked to state whether it is inside the circle, outside the circle, or on the circumference, then all we need to do is find the distance of that point from the centre of the circle....
    thanks very much, your method is differnet to 0- )'s right (can I use his/hers too)?

    also, if i'm given the equation of a circle, and it has = 4 at the end for example.. does that mean the radius is 4, or it is 4^2 ?

    Thanks
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    (Original post by Darkest Knight)
    thanks very much, your method is differnet to 0- )'s right (can I use his/hers too)?

    also, if i'm given the equation of a circle, and it has = 4 at the end for example.. does that mean the radius is 4, or it is 4^2 ?

    Thanks
    its the same method really. not much different, just more general, as it works for any circle.

    if it has 4 at the end, then r= \sqrt{4} = 2

    (well it depends what form it's in. perhaps post the equation here, in case it's in a different form..)
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    (Original post by GHOSH-5)
    its the same method really. not much different, just more general, as it works for any circle.

    if it has 4 at the end, then r= \sqrt{4} = 2

    (well it depends what form it's in. perhaps post the equation here, in case it's in a different form..)
    thanks, equaton is

    (x+3)^2 + (y-1)^2 = 4
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    (Original post by Darkest Knight)
    thanks, equaton is

    (x+3)^2 + (y-1)^2 = 4
    yes that's in the right form.

    if it isn't in that form, for example it's in the form:

     x^2+ y^2 - 2x + 4y -5 = 0

    then complete the square for x and y separately to get it in the right form...
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    (Original post by GHOSH-5)
    yes that's in the right form.

    if it isn't in that form, for example it's in the form:

     x^2+ y^2 - 2x + 4y -5 = 0

    then complete the square for x and y separately to get it in the right form...
    Thanks, so in that particular question... with the point being (-2,3). As 13>2 , it is outside of the circle?

    I'd appreciate if you could also help me on another question too lol.

    Equation of a circle is: x^2 + y^2 + 6x -4y +4 = 0, I need to prove that the line y = x + 5 passes through it.. and have no idea how!

    Thanks
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    (Original post by Darkest Knight)
    Thanks, so in that particular question... with the point being (-2,3). As 13>2 , it is outside of the circle?

    I'd appreciate if you could also help me on another question too lol.

    Equation of a circle is: x^2 + y^2 + 6x -4y +4 = 0, I need to prove that the line y = x + 5 passes through it.. and have no idea how!

    Thanks
    think simultaneous equations. if it passes through it, there will be two points where they intersect therefore if you sub y=x+5 into the eqn of the circle, you should get two solutions for x... (as the line is not vertical...)
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    (Original post by GHOSH-5)
    think simultaneous equations. if it passes through it, there will be two points where they intersect therefore if you sub y=x+5 into the eqn of the circle, you should get two solutions for x... (as the line is not vertical...)
    not sure if I did that right.. but I end p with x^2 + 6x +0.5 = 0 .. is that right?!

    Thanks
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    (Original post by Darkest Knight)
    not sure if I did that right.. but I end p with x^2 + 6x +0.5 = 0 .. is that right?!

    Thanks
    i dont think so, i could be wrong. post your working?
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    (Original post by GHOSH-5)
    i dont think so, i could be wrong. post your working?
    okay, the equation of circle, x^2 + y^2 + 6x -4y +4 = 0, after completing the square is (x+3)^2 + (y-2)^2 = 17
    Expand to get 2x^2 + 12x + 18 = 17
    divide by 2: x^2 + 6x + 9 = 8.5
    simplify x^2 + 6x + 0.5 = 0 ?

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