Arayan01
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Anyone know how to solve this? I found it in this paper: http://www.m4ths.com/uploads/3/2/7/4/3274186/358.pdf

Using algebra, show that part of the line 3x + 4y = 0 is a diameter of the circle
with equation x^2 + y^2 = 25.

So far, I managed to convert 3x + 4y = 0 into y = -3/4x + 0... and that's pretty much it, lol.
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RDKGames
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(Original post by Arayan01)
Anyone know how to solve this? I found it in this paper: http://www.m4ths.com/uploads/3/2/7/4/3274186/358.pdf

Using algebra, show that part of the line 3x + 4y = 0 is a diameter of the circle
with equation x^2 + y^2 = 25.

So far, I managed to convert 3x + 4y = 0 into y = -3/4x + 0... and that's pretty much it, lol.
If a segment of that line is the diameter then the line goes through the centre of the circle
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Arayan01
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(Original post by RDKGames)
If a segment of that line is the diameter then the line goes through the centre of the circle
But it says using algebra, and I'm pretty confused if we're allowed to show it through a diagram or not. Thanks for the reply btw
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RDKGames
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(Original post by Arayan01)
But it says using algebra, and I'm pretty confused if we're allowed to show it through a diagram or not. Thanks for the reply btw
Yes, you use algebra to show it analytically.

Substitute y=-\frac{3}{4}x into the circle and solve for x. Then find the coordinates of intersection between the line and the circle, and then show the distance between these two points is twice the radius.
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Arayan01
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(Original post by RDKGames)
Yes, you use algebra to show it analytically.

Substitute y=-\frac{3}{4}x into the circle and solve for x. Then find the coordinates of intersection between the line and the circle, and then show the distance between these two points is twice the radius.
OHHHHHHHHHHHHHHHHHHHHH! Legend! Absolute legend! I understand now, thanks mate!
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olivergeorge
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I would substitute y = -3/4x into x^2 + y^2 = 25 then solve for x (quadratic so should get two values of x). Substitute these values of x into y = -3/4x to find the co-ordinates where the straight line crosses the circle. Then since x^2 + y^2 = r^2, the diameter must be 10? Do a pythagoras with the two co-ordinates you found before and you should find that length is 10 - showing that it is a diameter.
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45.46 Litre Hat
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This is far simpler.

You already know the circle's centre, (0,0) using the equation, just show that the line passes through it using a substitution.
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RDKGames
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(Original post by 45.46 Litre Hat)
This is far simpler.

You already know the circle's centre, (0,0) using the equation, just show that the line passes through it using a substitution.
That's what I would've suggested in the first place but that doesn't involve algebra work as the question asks.
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45.46 Litre Hat
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(Original post by RDKGames)
That's what I would've suggested in the first place but that doesn't involve algebra work as the question asks.
It is completely algebraic - you know the centre from the circle's equation: (x-a)^{2}+(y-b)^{2}=r^{2}, where (a,b) is your circle's centre.

Also, there is no need to find a diameter's length.
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RDKGames
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(Original post by 45.46 Litre Hat)
It is completely algebraic - you know the centre from the circle's equation: (x-a)^{2}+(y-b)^{2}=r^{2}, where (a,b) is your circle's centre.

Also, there is no need to find a diameter's length.
I doubt that's what they're looking for, one liners aren't worth 6 marks and a full page.
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45.46 Litre Hat
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(Original post by RDKGames)
I doubt that's what they're looking for, one liners aren't worth 6 marks and a full page.
I really do agree, but I don't see what else you need to add.

Edit2: Edit: I realise the paper is a GCSE level test, of which the circle equation won't be part of the syllabus. no, the circle equation is on the syllabus, i'm really confused about why it's 6mks haha
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tiny hobbit
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(Original post by RDKGames)
I doubt that's what they're looking for, one liners aren't worth 6 marks and a full page.
(Original post by 45.46 Litre Hat)
I really do agree, but I don't see what else you need to add.

Edit2: Edit: I realise the paper is a GCSE level test, of which the circle equation won't be part of the syllabus. no, the circle equation is on the syllabus, i'm really confused about why it's 6mks haha
Going a long way round to prove something that can be done in a couple of lines doesn't make it better Maths. "Using algebra" doesn't really give much clue as to how much algebra is expected. I agree that 6 marks is far too many for the effort involved in showing that the line goes through (0,0). Maybe the man who wrote this practice paper hadn't thought of the quick way (apologies to him if I've missed something).

My model solution, putting in as much as I can:
The centre of the circle is (0,0). Substituting (0,0) into the LHS of the line equation gives 3x0 + 4x0 which equals 0 therefore the line passes through (0,0). Therefore part of the line is a diameter of the circle.
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fayeelw
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y = -3/4x
x^2 + (-3/4x)^2 = 25
x^2 + 9/16x^2 = 25
25/16 x^2 = 25
x^2 = 16
x = 4, y = -3
x = -4, y = 3

x^2 + y^2 = 25 = r^2 (circle rule)
therefore d = 10
line segment length = √6^2 + 8^2
= 10
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