Solving loci geometrically

[tall bough]

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[dawn nebula]

this is how i drew it centre (-7,4), but

you can also draw it like the red triangle here aswell and the angle would be minus pie/6. does it matter how i draw it?

also am confusd about the angle being negative or positive, do i use negative even though it is above real axis here but the half line is a negative angle

[sour thicket]

How I would solve for z is as follows: First off replace z with x+yi https://www.desmos.com/calculator/ofqrtijkbj Obviously you wouldn't have the graph, but the algebra is simple enough, then you find y using the first equation, and see which x+yi will give you an argument of -5pi/6.

as long as you can solve it one way, you don't need to know how to solve it in every possible way

Are you allowed to use a TI-84 calculator on this problem?

Cause that can handle complex numbers, you can't graph them but you can still plug them into the equations

[dawn nebula]

yeah true ig

not sure what calculator that is

i have a cg-100 from casio

i was just wondering because the geometric way the guy said would be faster to do

[sour thicket]

what did you get for z? just checking

[dawn nebula]

ill check

bit in blue

https://www.youtube.com/watch?v=sMzg6mj6Qdk&embeds_referring_euri=https%3A%2F%2Fsites.google.com%2F&source_ve_path=MjM4NTE this was the video he did it by chagning the equations into cartesian and solving simultanously

[sour thicket]

yes, that's what I got

wait, that equation on the top in blue -5pi/6 - pi/2 = -1/3pi isn't true. I think you meant +pi/2, not -pi/2?

how did you know the hypotenuse was 7?

[dawn nebula]

sorry went to toilet

the circle has modulus radius of 7

let me check that pie thing

ye i think i meant that but the angle is still -1/3 pie idk y i wrote thta

so i still use the negative angle i dont need to make it positive, cuz the half line angle is negative right

[sour thicket]

oh you're graphing the equations, I initially thought you were plotting down z on the complex plane

ok yea that makes a lot more sense

well, I wouldn't bother labeling the angle negative

i had never thought finding the intersection of a line and circle using this method tbh

I always solved for y, and substituted it in for y in the circle equation

but yes, that does seem a lot more efficient especially if you are given the angle of the line rather than the equation

[sour thicket]

oh wait, this only works because you know that the line passes through the center of the circle

[tall bough]

@dawn nebula

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[dawn nebula]

Oh ye true

[tall bough]

@sour thicket The user still needs help with this help request.

[dawn nebula]

Fuck

No I dont