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# Maths Ocr 2016 C1

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1. can someone explain 10 iii
2. (Original post by Logic4Life)
But I subbed in (10 0) for the second tangent on the next question!
The second part (second tangent) you had to find a new point where gradient also equals 2
3. Does anyone know the actual wording of all of q10? I swear there was a question asking to find the other tangent with a gradient of 2?
4. (Original post by 09brewh)
The second part (second tangent) you had to find a new point where gradient also equals 2
and how do u do that?
5. (Original post by Luketidley)
Circles also do not come up on C1, geometry of a circle is a topic on c2! The topics for C1 are;
Surds and Indices,
Quadratic functions, equations, graphs and transformations,
Coordinate geometry and straight lines,
Polynomials and the binomial expansion,
and Differentiation!

These are on my course and I'm doing WJEC.
Spoiler:
Show
and such confidence...
6. I had such high hopes as well 😂😂lol
7. (Original post by Misunderstood06)
and how do u do that?
let the gradient function: f(x) or dy/dx = 2... then solve for all values of x, cancel the ones already used
8. (Original post by Logic4Life)
But I subbed in (10 0) for the second tangent on the next question!
(Original post by fjfb1)
Does anyone know the actual wording of all of q10? I swear there was a question asking to find the other tangent with a gradient of 2?
For the first part it asked you for the tangent at point A to the circle.
You work out A through rearranging the circle formula (10,0) and then plug this into your line equation as this is the point your tangent you're looking for crosses.

I.e y =mx +c
therefore, 0 = 2(10) + c
therefore 0 = 20 + c
therefore c = -20

so the equation of the tangent was y = 2x -20, (since you've already worked out the gradient - 2).

For the second part it asked you for the tangent on the circle which was parallel to the first tangent (y= 2x-20). Obviously this would have to be on the other side of the circle, since they have to have the same gradient. So m = 2 again.
We know A (10,0), and we know Centre (4,3) so we can realise that the difference between the points makes up the radius, so add this difference again onto the centre and you will find the other side of the circle since it would be the length of the diameter, lets call this P.

Difference in x between A and C = -6, so x value of P, = 4(x value of Centre)-6 = -2. Difference in y between A and C = +3, therefore y value of P = 3+3 = 6. So coordinates of point P = (-2,6).

We know the gradient of the second tangent (2) since it is parallel to the first, and we now know the point it passes through, being (-2,6).
so, y= mx+c
therefore, 6 = 2(-2) + c,
therefore 6 = -4 +c
So 10 =c
As such, putting things together, the equation of the second tangent was y = 2x +10
9. (Original post by Misunderstood06)
and how do u do that?
Take the centre of the circle as the midpoint between the first and second point of same gradient
10. i seriously think i got 3 marks on that entire paper!!!!!!!!!!!!!
11. (Original post by Will487)
For the first part it asked you for the tangent at point A to the circle.
You work out A through rearranging the circle formula (10,0) and then plug this into your line equation as this is the point your tangent you're looking for crosses.

I.e y =mx +c
therefore, 0 = 2(10) + c
therefore 0 = 20 + c
therefore c = -20

so the equation of the tangent was y = 2x -20, (since you've already worked out the gradient - 2).

For the second part it asked you for the tangent on the circle which was parallel to the first tangent (y= 2x-20). Obviously this would have to be on the other side of the circle, since they have to have the same gradient. So m = 2 again.
We know A (10,0), and we know Centre (4,3) so we can realise that the difference between the points makes up the radius, so add this difference again onto the centre and you will find the other side of the circle since it would be the length of the diameter, lets call this P.

Difference in x between A and C = -6, so x value of P, = 4(x value of Centre)-6 = -2. Difference in y between A and C = +3, therefore y value of P = 3+3 = 6. So coordinates of point P = (-2,6).

We know the gradient of the second tangent (2) since it is parallel to the first, and we now know the point it passes through, being (-2,6).
so, y= mx+c
therefore, 6 = 2(-2) + c,
therefore 6 = -4 +c
So 10 =c
As such, putting things together, the equation of the second tangent was y = 2x +10
How many marks was that?
12. how many ums would 63 marks be.....hard to say but any guestimations??
13. i seriously think i got 3 marks on that entire paper!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!
14. (Original post by Logic4Life)
How many marks was that?
I believe the first part was 3 marks, and the second 2.
15. (Original post by alam13)
how many ums would 63 marks be.....hard to say but any guestimations??
seeming that the boundary for an A, going by the hardest of the other C1 papers, will be around 54/55, then i'd say you'd be comfortably in the 80s or even in the 90s for UMS
16. (Original post by AnonUK21)
Any ideas what the grade boundaries will be?

I got about 55 on that, which is terrible performance (~75%) compared to the 90ish%'s I was getting whilst doing past papers. Not sure if I just choked under pressure or if the exam was generally harder...
I'd imagine it to be 50-52 imo for an A.
It was harder than jan 11, which was 54 for an A.
17. Looking at the unofficial mark scheme I think I got around 59-61 depending on method marks. What UMS do you think this could be?

Does anyone have the paper?
18. What do people think 54-57 out of 72 will be UMS and grade wise. I really should have done better.
19. (Original post by Ozil5)
What do people think 54-57 out of 72 will be UMS and grade wise. I really should have done better.
A.

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