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Maths A level question

Hi,

There is this question which I got all the answers for but i wasn't sure about part C. I am just confused as to why the solutions of the equation is the answer for finding the coords of B and C and not just the roots of the equation.

Thanks in advance for help
Reply 1
Screenshot 2023-05-30 210324.png
How did you do part b, if you don't mind me asking?
Reply 3
Original post by toxicgamage56
How did you do part b, if you don't mind me asking?


It's quite a tricky one and I got a bit lucky tbh seeing it like this quite quick.

You have to imagine it as a circle with A as the centre because we know that 2 radii to the edge of a circle make an isosceles triangle, with the B and C being the points to the edge of a circle. Then using 2 root 13 as the radius and the equation of a circle, you can expand and get the quadratic it gives you.
(edited 11 months ago)
Original post by pinkrocket
It's quite a tricky one and I got a bit lucky tbh seeing it like this quite quick.

You have to imagine it as a circle with A as the centre because we know that 2 radii to the edge of a circle make an isosceles triangle, with the B and C being the points to the edge of a circle. Then using 2 root 13 as the radius and the equation of a circle, you can expand and get the quadratic it gives you.

Ah, I get it. That's super smart, it's scary to think that they expect us to look at the question through the angle despite no mention of the circle lol. Thank you for pointing that out.
Reply 5
Original post by toxicgamage56
Ah, I get it. That's super smart, it's scary to think that they expect us to look at the question through the angle despite no mention of the circle lol. Thank you for pointing that out.


I know its such a weird thing for them to ask. I'm glad I got it but I doubt i'd do something like that in the real exam haha
Original post by pinkrocket
I know its such a weird thing for them to ask. I'm glad I got it but I doubt i'd do something like that in the real exam haha

Yeah, hopefully they don't. Btw, for c, isn't the coordinate for B (-8/5, -2/5)?
Reply 7
Original post by toxicgamage56
Yeah, hopefully they don't. Btw, for c, isn't the coordinate for B (-8/5, -2/5)?


I assume you mean (-8/5,-21/5). Yes, that's the right answer I was just confused because to get that you have to solve the quadratic but I thought that just got you to the roots and wouldn't get you to the co-ordinates of B or C.
Original post by pinkrocket
I assume you mean (-8/5,-21/5). Yes, that's the right answer I was just confused because to get that you have to solve the quadratic but I thought that just got you to the roots and wouldn't get you to the co-ordinates of B or C.

No the quadratic is what you get after subbing the equation of the line into the circle equation so by solving that quadratic, what you get are intersection points between the circle and line not roots.
Reply 9
Original post by toxicgamage56
No the quadratic is what you get after subbing the equation of the line into the circle equation so by solving that quadratic, what you get are intersection points between the circle and line not roots.


ahhhh, yes thank you so much. Funny how i can get the hard bit and not the more simple bit lol
Original post by pinkrocket
ahhhh, yes thank you so much. Funny how i can get the hard bit and not the more simple bit lol

No problem. I came here trying to help, but instead I got stuck on part b and ended up needing help myself lol, glad I could still be of assistance.
Reply 11
Original post by pinkrocket
Hi,

There is this question which I got all the answers for but i wasn't sure about part C. I am just confused as to why the solutions of the equation is the answer for finding the coords of B and C and not just the roots of the equation.

Thanks in advance for help


Youve obviously sorted the numbers/realised that its just a line intersecting a circle so a gcse++ question/solution (not clearly stated and the circle not at the origin), but it got me thinking about what it means when you substitute the line into the circle equation and solve the quadratic in x (find the roots) as obviously its correct but its handle turning algebra. In hindsight its fairly obvious but ..

So to take a step back, a point on l1 is given by (x, 2x-1) and you want the distance from (-2,3) to be 2sqrt(13). The most basic thing to do is find the distance (squared) using pythagoras on the difference of the coordinates
d^2 = (x+2)^2 + (2x-4))^2
Obviously this is similar to your circle equation (a circle is the locus of points which are equidistant from the centre, a basic definition which is easy to forget, so the circle equation is just pythagoras), the circle just has
d^2 = r^2
which in this case r^2 is 52.

But the basic quadratic
d^2 = 5x^2 - 12x + 20
measures the distances of points on l1 from (-2,3). So for instance we could complete the square to get
d^2 = 5(x-6/5)^2 + 64/5
which tells us that the closest approach (intersection between l1 and l2) occurs at the point (6/5,7/5) and the distance is 8/sqrt(5).

Returning to the question, you want to solve
d^2 = 52
which correpsonds to the root finding problem of
d^2 - 52 = 5x^2 - 12x - 32 = 0

So the two points on l1 corresponding to the two roots of the quadratic are exactly a distance of 2sqrt(13) from (-2,3) and the other values of the quadratic correspond to the distance (squared and -52) from (-2,3). As its a "u" quadratic, it will be negative in the centre (d^2<52) corresponding to points on the chord inside the circle (the chord is inside so the distance must be less than the radius). Also, fairly obviously, the quadratic is positive when d^2>52 so the points are on the chord extended outside the circle.

Thinking of it as pythagoras/distance explains why the quadratic roots "works", though its no different from line intersecting circle, sub for y, find the roots algebra. In hindsight it probably falls into the "bleeding obvious" category, but its easy to miss just handle turning the algebra and while its a good spot to think of it as line intersecting circle, you dont need to as you could simply calculate the distance between points (pythagoras) without making that link.
(edited 11 months ago)
Reply 12
That a lot of typeing

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