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Edexcel question (new spec)

How would you answer this question?
A is the centre of circle C, with equation x^2- 8x + y^2 10y + 1=0
P, Q and R are points on the circle and the lines L1, L2 and L3 are tangents to the circle at these points respectively. Line L2 intersects line L1 at B and line L3 at D.
a) Find the centre and radius of C. (3 marks)
b) Given that the x-coordinate of Q is 10 and that the gradient of AQ is positive,find the y-coordinate of Q, explaining your solution. (4 marks)
c) Find the equation of l2, giving your answer in the form y = mx b. (4 marks)
d) Given that APBQ is a square, find the equation of L1 in the form y = mx b. (4 marks)
L1 intercepts the y-axis at E.
e) Find the area of triangle EPA. (4 marks)

Reply 1

vc94

(a) Use completing the square twice in order to write the equation in the form:
(x-a)^2 +(y-b)^2 = r^2

Reply 2

Flukeyboy

Original post by vc94

(a) Use completing the square twice in order to write the equation in the form:
(x-a)^2 +(y-b)^2 = r^2

Yes I have completed part a but part b onward I need help

Reply 3

CraigFowler

Original post by Flukeyboy

Have you drawn a sketch? For part 2, plug in the x=10 into the equation you developed for the circle to obtain y value. Part 3 Gradient of L2 will be perpendicular to AQ, so find negative reciprocal and substitute in the coordinates of Q for the full equation. Similar for rest of question

Reply 4

Flukeyboy

Original post by CraigFowler

yh thanks. i got the answer

Reply 5

rachela24

Hey, what did you get for d?

Reply 6

Lol12309873645

Very funny, btw you look like a white version of my big toe.

(edited 4 years ago)

Reply 7

Josh burns

Original post by rachela24

Hey, what did you get for d?

....

Reply 8

JoEmomm

How did you find the equation of L1?

Reply 9

-spaceb0y

I came to this thread for the answer and was dismayed that I couldn't find it.
So I worked it out.
Here is the answer for part d):
The gradient of l1 is 1/3 (negative reciprocal of the gradient of l2 (perpendicular to l1), which is -3)
Substitute y = 1/3x + b into y in the circle equation (x-4)^2 + (y+5)^2 = 40 and expand fully.
You should get 10x^2 + (6b - 42)x + 9b^2 + 90b +9
The line is a tangent to the circle so the discriminant will equal 0 (b^2 - 4ac = 0).
Find the discriminant on its own.
Expanded and made equal to 0, this will give you 324b^2 + 4104b -156 which simplifies to 3b^2 + 38b -13 = 0
Factorise this to get (3b - 1)(b + 13)
Therefore b = 1/3 or -13
We can see the intersect of l1 is positive, so we can rule out -13.
This leaves us with the equation of l1 being:

y = 1/3x + 1/3

Part e is easy when you know this.
I hope this helps many unfortunate people who have been set this question to do by their maths teachers.

(edited 3 years ago)

Reply 10

DFranklin

Original post by -spaceb0y

I came to this thread for the answer and was dismayed that I couldn't find it.

I hope this helps many unfortunate people who have been set this question to do by their maths teachers.

I hate to rain on your parade, but (unless I'm very mistaken) you can answer this much more simply if you think a little more geometrically rather than diving straight into algebra.

Looking at the marks for each part I'm sure you weren't supposed to be using a discriminant solution, but all respect to you for pushing that solution through.

In the case of a 2 year old thread no-one's really going to care, but please bear in mind there's a general rule against posting full solutions.

Reply 11

SonyaC

Original post by DFranklin

What would the easier solution be?

Reply 12

DFranklin

Original post by SonyaC

What would the easier solution be?

Just use that ABPQ is a square, and that for any point X on the circle, the tangent at X is perpendicular to AX.

Reply 13

SafwanAli

But you don’t have any coordinates so that doesn’t get you any closer to the answer unfortunately

Reply 14

DFranklin

Original post by SafwanAli

But you don’t have any coordinates so that doesn’t get you any closer to the answer unfortunately

You have the coordinates (from earlier parts) of A and Q.

Reply 15

Blueberry Jam

Original post by DFranklin

You have the coordinates (from earlier parts) of A and Q.

Hello, I am having trouble with the question still (part d).
How do I solve this 'geometrically' as you previously mentioned? I understand that the equation of l1 has m=1/3, but how do I move forward without any points on l1?

Reply 16

TazSheffield

You have the coordinates of point A and the gradient of the line AP is the same as the gradient of l2 as they are parallel so work out the equation of AP and then solve this equation with your circle equation to work out the coordinates of point P, so use this point with the gradient of l1 to work out your required equation. I hope this helps if you are still struggling with it.
Even more easier than that using the fact that APBQ is square you can work out the coordinates of point B by looking at how you get to Q from A (I mean across and up) and then use the gradient and point B to get your equation.