Graphing question

[cold frigate]

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[white flame]

Find the points, then consider what length the chord of that angle has to be.

[elder hare]

how can i find the points

i know they lie on the circle

that has a radius of 1 from origin

[white flame]

Express x or y from the line equation, substitute into the circle equation. That gives a quadratic equation.

[elder hare]

x^2+(x+m)^2=1
2x^2 + 2xm + m ^2 = 1

can i sub 1 in for x?

[white flame]

No. Solve for x.

[elder hare]

how do i factorise 2x^2 + 2xm + m ^2 = 1 when there a m^2 at the end

[white flame]

You can't factor it, just solve for x.

[minor dagger]

It’s easier to find the horizontal distance from A to B and note that’s the difference of the roots of the quadratic

Then do Vieta’s

[elder hare]

i havnt learnt vieta yet

[white flame]

Oh, right, good idea! Since (x2 - x1)^2 = (x2 + x1)^2 - 4x2 x1.

[white flame]

I doubt that.
Suppose x1 and x2 are the roots of ax^2 + bx + c. Then:
x1 + x2 = -b/a
x1 x2 = c/a

[elder hare]

ok i get it

[proven spruce]

I think theres a simpler way to see it

[white flame]

Oh, I have another idea.

[proven spruce]

since the equation of a line is always 45º, and the angle between the origin and the 2 intersections is 60º, this means its a side of a hexagon

[white flame]

Good idea, but it's too much of a particular approach for me.

[proven spruce]

if t is the length of a side of a hexagon, then m = t/2 (nvm I think its wrong)

[white flame]

Let's pick two points on a circle: (cos(t1), sin(t1)) and (cos(t2), sin(t2)). Then:
t2 - t1 = π/3
sin(t1) - cos(t1) = m
sin(t2) - cos(t2) = m