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    (Original post by jordanwu)
    I got root 34 for the radius as well.. I'm so confused.. The centre of the circle was given as (5,-3) so (x-5)^2 + (y+3)^2 = 25+9 .....
    Hmmmm i'm not sure which is the correct answer, but after finding the length of AC, I've got root 65 as A is on the circumference of the circle, so by finding the length from the centre to the circumference, we're actually calculating the diameter, which was root65, but the final answer is just =65 (RHS)

    hope it helped
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    (Original post by tajtsracc)
    I'm pretty certain radius is root 65. Then for the last question, you use pythag, 65 + 16 = 81, square root of that gives you a solid 9.
    no i got root 34
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    (Original post by Poppy1098)
    Exactly!! Because -25-9 is -34??
    Also what did you get for CT?
    You work out k (or r^2) by finding the distance of CA squared, which was (-5-2)^2 + (3- -1)^2 = 7^2 + 9-4)^2 =65

    I got 9 for CT by doing (√65)^2 + 4^2 = 81, then √81 = 9
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    (Original post by melanin101)
    How are you guys getting root 34 for the radius?? I got root 65 by using Pythagoras to find the length of the centre to the point they gave us...

    root 34 people, what did you get for the length CT?
    I realised I got it wrong radius root34
    CT question I put root50
    hope they give us marks for method
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    (Original post by WhiteTigerKURD)
    I got 11.25
    I got the same I hope we are right !!! Mine was in fraction form like 135/12
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    (Original post by Parhomus)
    The method you talk about is when you have something like ax^2+bx+cy^2+dy=k
    I almost did the same thing but then when I plugged in coordinates and realised it was wrong I saw you had to find the radius from A and C and it gave you sqrt65.
    Fml, imo this paper was harder than last year's because there were so many questions where you could make mistakes on. Do you remember what the coordinates of A were?
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    (Original post by yasaminO_o)
    You work out k (or r^2) by finding the distance of CA squared, which was (-5-2)^2 + (3- -1)^2 = 7^2 + 9-4)^2 =65

    I got 9 for CT by doing (√65)^2 + 4^2 = 81, then √81 = 9
    Ahh thank you, that makes sense! Will I get an error carried forward for the CT question then?
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    I got 9.


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    (Original post by moe889)
    no i got root 34
    I'm sure it has to be root 65. What was your CT? I got √81 = 9, which can't be a coincidence.
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    Can you still get marks on the integration if the answers was wrong?? Like method marks
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    (Original post by Jacobisswaggy)
    For the coordinates of Q, I don't think I wrote (-5/4,0) I just wrote -5/4 - did you have to give the full coordinate? It was obvious that y was 0 :/
    only had to give the x value
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    (Original post by haza01)
    thats exactly what i did
    Yeah same except it came out as k2-24k+80 so (k-20)(k-4) giving k=20 and k=4

    anyone else get that?
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    (Original post by Poppy1098)
    Ahh thank you, that makes sense! Will I get an error carried forward for the CT question then?
    From looking at past mark schemes I'm sure you'll get the method marks but you probably won't get the final answer mark
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    (Original post by tajtsracc)
    I'm sure it has to be root 65. What was your CT? I got √81 = 9, which can't be a coincidence.
    i think youre wrong mate. i got 9 aswell
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    its all ogre now
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    (Original post by kylabaebae)
    Yeah same except it came out as k2-24k+80 so (k-20)(k-4) giving k=20 and k=4

    anyone else get that?
    I got that
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    (Original post by hlincoln)
    Can you still get marks on the integration if the answers was wrong?? Like method marks
    Yeah I messed up aswell. But luckily I got the area of the triangle right, 9, which is one mark (looking at past papers). One mark for integration and one mark for substituting x into the integrated formula. I mentioned "intergral - area of triangle" which I'm not sure counts, but if it does then that's 4/8. Not too bad.
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    is there an unofficial mark scheme up yet?
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    (Original post by tajtsracc)
    Yeah I messed up aswell. But luckily I got the area of the triangle right, 9, which is one mark (looking at past papers). One mark for integration and one mark for substituting x into the integrated formula. I mentioned "intergral - area of triangle" which I'm not sure counts, but if it does then that's 4/8. Not too bad.
    Which Year and month of past year paper are you referring to?
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    (Original post by Pingu7)
    Igot -0.5 because I sketched the curves and it moved left
    yeah I did the same and also got -0.5...?
 
 
 
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