circle equation

[idle cobalt]

Find the equation of the tangents from point A(1;0) to circle k:(x-3)^2 + (y-1)^2 = 1

[thorny temple]

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[winter remnant]

What have you tried?

[idle cobalt]

not sure how should I go about it

[winter remnant]

You want the tangent, right? What do you pretty much always have to do when you want the tangent?

[idle cobalt]

If a line is tangent to a given circle, it is perpendicular to the radius at the point of contact.

[winter remnant]

...that's... not a thing for you to do.

[idle cobalt]

sorry

[winter remnant]

We need the equations for the lines, right? What do we need to know for the equation of a line?

[idle cobalt]

https://cdn.discordapp.com/attachments/949615984120041495/1140637171284451398/image.png

[winter remnant]

...we need to know the general formula for the equation of a circle to write the equation of a line?

[idle cobalt]

the equation of a line is Ax + By + C = 0

@runic ember

[winter remnant]

...what two pieces of information about a line do we need to know in order to write its equation?

[idle cobalt]

x and y

[winter remnant]

You're guessing. Stop guessing.

[idle cobalt]

I'm answering the question

[winter remnant]

You're answering the question so incorrectly that it indicates that you don't really understand it and are therefore guessing.

[idle cobalt]

so what should I do instead

[winter remnant]

...well, let's think about it. The equation of a line can also be expressed as y = mx + b, right?

[idle cobalt]

so it's m and b

[winter remnant]

...which are what?

[idle cobalt]

the two pieces of information about a line which we need to know in order to write its equation

[winter remnant]

...the slope. We need the slope and a point.

[idle cobalt]

we have a point

[winter remnant]

And we're told something about the slope, aren't we?

[idle cobalt]

let me check

it should be tangent

[winter remnant]

And how do we find the slope of a line tangent to a curve?

[idle cobalt]

I haven't done calculus

[winter remnant]

Who said anything about calculus?

[idle cobalt]

there is something about derivatives which is a calculus topic

[winter remnant]

"There is something about derivatives" where?

[idle cobalt]

to find the slope of a line tangent to a curve in specific point

[winter remnant]

But where though? Where are you getting this information?

[idle cobalt]

I think we need to find the derivative for that

google

[winter remnant]

...I really don't know how you're supposed to solve this problem without calculus.

[idle cobalt]

we haven't done calculus in school

[dry parrot]

Take the tangent perpendicular to the radius

[winter remnant]

Which radius?

[dry parrot]

The radius through A, the tangent is perpendicular to that

[idle cobalt]

I think they are refering to this

[winter remnant]

...no, no it's not? The tangent perpendicular to a radius that passes through A does not pass through A.

[dry parrot]

Doesn't it say the tangent at A

Ignore me I thought A was on the circle lol

But we can say the equation of a line through A has form y=mx-m

And we want 1 intersection with the circle

[winter remnant]

...can we say that?

[dry parrot]

So substitute the equation of line into the circle, the resulting quadratic has a discriminant of 0

[dry parrot]

m(1)-m=0, so (1,0) lies on it, right?

[winter remnant]

That is one line through A. It's not all lines through A.

[idle cobalt]

analytic geometry is so difficult

[dry parrot]

If you let the line have slope m, then using point slope, $$y-0=m(x-1) \implies y=mx-m$$

[delicate briar]

Civil Service Pigeon

[winter remnant]

...huh. Okay then.

[idle cobalt]

I didn't know I will need calculus for that

[dry parrot]

You don't

[idle cobalt]

[dry parrot]

Does point-gradient form sound more familiar

Some ppl use gradient and some ppl use slope

[idle cobalt]

I use neither (as a non native speaker)

[dry parrot]

Do you guys understand or no

[idle cobalt]

I still don't know how to solve it

[winter remnant]

...yeah, I understand. That's why I said "Okay then."

[dry parrot]

Ah ok

[winter remnant]

Are we sure though that the discriminant should be 0? I think maybe it should be positive, since there's two tangents.

[dry parrot]

The ... huh just made me think of when ppl give up and just accept a result without understanding it

[dry parrot]

But each line intersects only once

[idle cobalt]

it should be one

[dry parrot]

A positive discriminant means 2 intersections with the line, which wouldn't be a tangent

[winter remnant]

Wait, no, I got it. The discriminant is a quadratic in m.

[dry parrot]

👍

Can you explain it for me

I gtg soon

[winter remnant]

So have you followed the explanation so far?

[idle cobalt]

i think I'm lost

[winter remnant]

Okay, where did we lose you?

[idle cobalt]

around the part with calculus

[winter remnant]

We didn't do any calculus.

[idle cobalt]

but we talked about it

[winter remnant]

You talked about it.

[idle cobalt]

this is where you lost me

[winter remnant]

So you basically missed Pigeon's entire explanation?

[idle cobalt]

yeah

[winter remnant]

...do you know the point slope form of a line?

[idle cobalt]

y − y1 = m(x − x1)

we need to know:
-one point on the line: (x1, y1)
-and the slope of the line: m

[winter remnant]

The point is (1, 0).

[idle cobalt]

but what is the slope

[winter remnant]

...we don't know that yet. That's what we're trying to find out.

[idle cobalt]

and how do we find out

[winter remnant]

BY PUTTING THE POINT (1, 0) INTO THE EQUATION.

[idle cobalt]

ok

please don't shout

I can hear you

[winter remnant]

We've been doing this for 40 minutes, I haven't had breakfast yet, and you're basically forcing me to repeat everything someone else already just said because you weren't paying attention the first time.

[idle cobalt]

y - 0 = m(x - 1)

[idle cobalt]

sorry I'm dyslexic

you can leave if you want

[winter remnant]

I don't see how that affects your ability to pay attention.

[idle cobalt]

I will wait for someone else to help

[winter remnant]

...look. I'll take a break, and I'll come back, alright. Sorry I got snippy.

[idle cobalt]

I don't want to make you feel forced

take your time

[winter remnant]

...okay, I feel better now. I really was just hungry.

[winter remnant]

So to start with, simplify this.

[vague swift]

Well, let's look at it generally.
For the slope of the tangent we need the derivative of y, so dy/dx. And we are given an implicit function F(x, y) = C.
How can we find dy/dx from the equation of the implicit function?

[idle cobalt]

y = mx - m

[winter remnant]

They haven't taken calculus.

[idle cobalt]

yes

[winter remnant]

And the circle is (x - 3)^2 + (y - 1)^2 = 1, right?

[vague swift]

Hm, that's odd... Not sure how you can work with tangents to functions without calculus. Generally, at least.

[idle cobalt]

correct

[winter remnant]

So we plug y = mx - m into (x - 3)^2 + (y - 1)^2 = 1.

And that'll get us the points at which the line and the circle intersect.

[vague swift]

Ohh. That's how you want to do it, I see. Clever!

[winter remnant]

It was Pigeon's idea. I'm just explaining it.

[vague swift]

Ah, alright.

[winter remnant]

@idle cobalt Can you expand (x - 3)^2 + (mx - m - 1)^2 = 1 and get everything on one side to get it into quadratic form? That is, ax^2 + bx + c = 0?

[idle cobalt]

let me see

we get

(1+m^2)x^2 + (-6-2m^2-2m)x + 9 + m^2 + 2m = 0

if my calculations are right

[winter remnant]

...I think it's supposed to be an 8, not a 9.

[idle cobalt]

1 and 1 cancel out

[winter remnant]

Yeah, I got it wrong.

[winter remnant]

So do you know what a discriminant is?

[idle cobalt]

b^2 - 4ac

[winter remnant]

So the equation that we have is for the points of intersection between y = mx - m and (x - 3)^2 + (y - 1)^2 = 1. Since we're looking for tangents, we want there to be exactly one such point, so we want to find the value(s) of m for which the quadratic we've derived has exactly one solution. What does that require?

[idle cobalt]

the discriminant must be equal to zero

[winter remnant]

Right. So calculate the discriminant of the derived quadratic, set it equal to 0, and solve for m.

[idle cobalt]

m1 = 0 and m2 = 4/3

[winter remnant]

Right.

So our tangent lines are what?

[idle cobalt]

if m = 0 then y = y1

y = y1 = 0

[winter remnant]

No. 1) they don't know calculus. 2) A isn't on the circle.

3. we already finished the entire problem.

[idle cobalt]

Thank you for the walkthrough. I am just extremely ashamed to admit that my intelligence quotient is below average so it's kinda difficult for me to comprehend complex(from my perspective) problems from first try.

Maybe I should have mentioned that earlier because most of the teachers I know are not comfortable wasting time with students like me and would rather avoid them.

[oak charm]

@tulip gate has given 1 rep to @winter remnant

[winter remnant]

No, man, it's totally not your fault. I didn't see this solution method either.

[idle cobalt]

I changed my about me to avoid future nuisance