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# AQA C1 Unofficial Markscheme

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1. (Original post by timtjtim)
The latter. See image.
2. (Original post by beanigger)
to work out the radius of circle u needed to find the "difference" in x and y directions and use Pythagoras to find the actual length of it
if you do M1 you would be very familiar with this.
think of it like this, how do you get from coordinates (10,10) to (5,6)
well, you go -5 in the x-direction and -4 in the y-direction
now think of x and y as being horizontal and vertical respectively.
what is the hypotenuse of that triangle u have? ( hypotenuse = radius of the circle)
Nice, see image for a graphical explanation of this. You could also substute A values into the (x-5)^2 (y+3)^2 = k to get k.
Attached Images

3. (Original post by AyyS)
Quick question, you know question 1. Did it give us the equation of AB or did we have to create it with the c=7. Or was the y=mx+7 there to confuse us??
y=mx+7 was there to confuse us. They just wanted the gradient (m). Of course, m is the same in both, but giving y=mx+7 was just messing with our heads
4. (Original post by Agent R5)
Then something else about values for k I think it was (or was it x?) And I got k>6 and k<-3/2.
I really can't remember, but basically it hahas a quadratic equation with values a, b and c in terms of k, and you had to take the discriminate of that and solve it for it to.....oh man I can't remember now and am beginning to question myself, i may have gotten that all wrong.

I'm sure i got that answer somewhere, if anyone can remember what the actual question was?
5. (Original post by timtjtim)
Nice, see image for a graphical explanation of this. You could also substute A values into the (x-5)^2 (y+3)^2 = k to get k.

What did you get the coordinates of B as?
6. i got root 34 for radius , woooooo !!!!! i dont believe that it can be anything else.....peasants.....
7. Hey, I can't remember all the questions but here are some of my answers. Let me know if you disagree with any!
1) m=-5/3 B(-3,4)k=-30
2) (3rt5)^2=45 75-32rt5
Can't remember the rest so here are some of my other answers I remembered in random order:
Area under curve=81/4 Area shaded region=45/4
(x-7/4)-41/4 (asked for rational form so use fractions) minimum point=-41/4 graph translated by vector [1/2, 41/4]
d^2y/dx^2=-2x-9x^2 sub in x-coord of p to get 45, 45>0 therefore minimum
B(12,-7) m=-4/7 m(tangent)=7/4 7x-4y+18=0
tangent for k (used b^2-4ac=0) k=20 k=4
(x-5)^2+(y+3)^2=65 length of CT=9
y=-32x-40 x-coord Q=-5/4
For graph I had n-shaped parabola intersect y-axis at 8 and x-axis at -1+-rt5
Three linear factors were (x-3)(x-4)^2
For the remainder theorem I think I got 20
8. (Original post by musslih)
I thought the equation for the tangent at P is y=-40x-56..... since Y-24=-40(x-2), how did you guys get gradient as -32...?
i got that too, we both must have made a mistake! which is annoying as that means we got Q was -4/7 and so the area was also wrong, how many marks will we loose?
9. (Original post by Cassette)
I didn't think it was that bad, none of the questions were worded in a confusing way which is what usually gets me. However, as others say, there were some big numbers and unexpected elements. Was the radius of that circle sqrt(65) or sqrt(69) or something like that? I forget the answer I gave, but several times like that I stopped and went back to check my working, which is what ultimately cost me the time I needed to finish. I should probably have just trusted myself, but in practice papers I learnt that if it feels like an unexpected answer, it's probably wrong. I was about a rushed five minutes away from complete - skipped over a 2 mark and a 3 mark sub-question, with a little working on the latter at the end. So at best I've already dropped four marks, because I was too slow. Not happy with myself.
This sums it up perfectly. Felt the questions were tough but not much differnce from last year. I've always taught myself that exam papers (non calculator ones anyway) are done so that you get nice simple numbers and so when I did the circle one and kept getting root 65 I thought i'd lost it mentally. Really wasted a lot of re-doing questions multiple times because of hard/weird the numbers were. I think grade boundaries will be just below last year's just as there was a lot of room for error with so many long dodgy calculations.
10. (Original post by timtjtim)
To find the circle given C, you don't complete the square. The centre was 5,-3. You can say those are (a,b).

Now plug a,b into (x-a)^2 + (y-b)^2 = r^2

We get (x-5)^2 + (y+3)^2 = r^2.

Now for r - we know that A lies on this circle and is at coordinates A(-2,1). That gives a triangle, with sides 4 and 7. 7^2 + 4^2 = 65, so r = √65. But we want r^2, and √65^2 = 65 so k = 65.

My image may help - see attached.
Didn't sit the exam so won't correct quoted numbers.

So we have

we know that there is a point on the radius that is located at (-2, 1) and that this point must therefore satisfy Eq. 1.

Therefore

so is correct though you really don't need to faff with the triangle stuff to work it out.
11. (Original post by money-for-all)
i got root 34 for radius , woooooo !!!!! i dont believe that it can be anything else.....peasants.....
See attached for why it was root 65.
Attached Images

12. (Original post by df97)
Hey, I can't remember all the questions but here are some of my answers. Let me know if you disagree with any!
1) m=-5/3 B(-3,4)k=-30
2) (3rt5)^2=45 75-32rt5
Can't remember the rest so here are some of my other answers I remembered in random order:
Area under curve=81/4 Area shaded region=45/4
(x-7/4)-41/4 (asked for rational form so use fractions) minimum point=-41/4 graph translated by vector [1/2, 41/4]
d^2y/dx^2=-2x-9x^2 sub in x-coord of p to get 45, 45>0 therefore minimum
B(12,-7) m=-4/7 m(tangent)=7/4 7x-4y+18=0
tangent for k (used b^2-4ac=0) k=20 k=4
(x-5)^2+(y+3)^2=65 length of CT=9
y=-32x-40 x-coord Q=-5/4
For graph I had n-shaped parabola intersect y-axis at 8 and x-axis at -1+-rt5
Three linear factors were (x-3)(x-4)^2
For the remainder theorem I think I got 20
well done for remembering all of that bruv, well done.
good for you.
go get em' tiger
ROAR !
excellent.
13. (Original post by moe889)
i did too but some guy called tajtsracc keeps saying its root 65
i got root 34 broooooda
14. (Original post by pb123)
yhi put that too!! its right just not simplified.... it goes to 81/4 so probz get 4 marks out of 5
It didn't ask for it in the simplest form so I think we should get full marks
15. hi, just wondering why the area is 9 as i did the area of the rhombus? :/
16. (Original post by timtjtim)
See attached for why it was root 65.
(x-a)^2 +(y-b)^2 = k
where k = r^2 = 65
why you lot are saying sqroot65 is beyond me?
17. (Original post by priyanka1402)
hi, just wondering why the area is 9 as I did the area of the rhombus? :/
It's a triangle.
Attached Images

18. (Original post by priyanka1402)
hi, just wondering why the area is 9 as i did the area of the rhombus? :/
AQA were being knobs you did not even need R(1,0) for the whole question, yet still stated it in the q?
19. (Original post by pb123)
yhi put that too!! its right just not simplified.... it goes to 81/4 so probz get 4 marks out of 5
It didn't say you needed to write it in the simplest form so we should get full marks
20. There was a surd question at the start of a new question (Q3 maybe) (not the surd one which was question 2 I think)
Which said put the X values for the quadratic in the form m plus or minus root n

I put the quadratic into a completed the square form then rearranged to get X=

I ended up with X = 1 plus or minus root 5

But my friend had -1 and I can't remember which will be right (+/-) as I can't remember the quadratic

What did you guys get

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