Flukeyboy
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#1
Report Thread starter 3 years ago
#1
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)
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vc94
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#2
Report 3 years ago
#2
(a) Use completing the square twice in order to write the equation in the form:
(x-a)^2 +(y-b)^2 = r^2
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Flukeyboy
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#3
Report Thread starter 3 years ago
#3
(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
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CraigFowler
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#4
Report 3 years ago
#4
(Original post by Flukeyboy)
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)
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
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Flukeyboy
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#5
Report Thread starter 3 years ago
#5
(Original post by CraigFowler)
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
yh thanks. i got the answer
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rachela24
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#6
Report 3 years ago
#6
Hey, what did you get for d?
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Lol12309873645
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#7
Report 2 years ago
#7
Very funny, btw you look like a white version of my big toe.
Last edited by Lol12309873645; 2 years ago
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Josh burns
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#8
Report 1 year ago
#8
(Original post by rachela24)
Hey, what did you get for d?
....
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JoEmomm
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#9
Report 1 year ago
#9
How did you find the equation of L1?
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-spaceb0y
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#10
Report 1 year ago
#10
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.
Last edited by -spaceb0y; 1 year ago
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DFranklin
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#11
Report 1 year ago
#11
(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.
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SonyaC
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#12
Report 1 year ago
#12
(Original post by DFranklin)
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.
What would the easier solution be?
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DFranklin
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#13
Report 1 year ago
#13
(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.
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SafwanAli
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#14
Report 1 year ago
#14
But you don’t have any coordinates so that doesn’t get you any closer to the answer unfortunately
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DFranklin
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#15
Report 1 year ago
#15
(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.
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Blueberry Jam
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#16
Report 11 months ago
#16
(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?
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TazSheffield
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#17
Report 11 months ago
#17
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.
Last edited by TazSheffield; 11 months ago
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Real.A
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#18
Report 10 months ago
#18
Hey I’m struggling in par b onwards
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rafas2003
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#19
Report 9 months ago
#19
Likewise
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