Maths Ocr 2016 C1

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

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.

and how do u do that?

But I subbed in (10 0) for the second tangent on the next question!

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?

and how do u do that?

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?

how many ums would 63 marks be.....hard to say but any guestimations??

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...

What do people think 54-57 out of 72 will be UMS and grade wise. I really should have done better.