#Determine a circles equation based on points in the perimeter, radius and base coordinate system

1 messages · Page 1 of 1 (latest)

charred badge
#

I am given this image with points A & B (C & D seems irrelevant for now), and with the radius being 25 I thought I could make a circle in point A & B with the same radius, then get the intersection to determine x and y of the middle circle to be able to determine the equation of the circle.

So i made these:
A: (x-3)^2+(y-38)^2=25^2
B: (x-26)^2+(y-25)^2=25^2

And I expanded & simplified

A:
x^2+2^2-6x+y^2+38^2-76y=25^2
Simplified A:
x^2+y^2-6x-76y+1448

B:
x^2+26^2-52x+y^2+25^2-50y=25^2
Simplified B:
x^2+y^2-52x-50y+1301

I subtract B from A and get this
46x-26y+147=0

Move Y alone
26y=46x+147
Divide by 26

26y/26=46/26 x+147/26

And then I have Y being
y=(23/13)x+147/26

I insert and I have this monstrosity
(x-3)^2+(23/13 x+147/26-38)^2-625=0

How do I expand and simplify this monster, and is it even the right approach

image.png
tough raptorBOT
#
  1. Wait patiently for a helper to come along.
  2. Once someone helps you, say thank you and close the thread with:
+close
  1. Feel free to nominate the person for helper of the week in #helper-nominations
  2. Do not ping the mods, unless someone is breaking the rules.
  3. If you're happy with the help you got here, and the server overall, you can contribute financially as well:
onyx torrent
charred badge
onyx torrent
#

Let me try.

onyx torrent
# charred badge I plotted it into geogebra and I solved for x in this: (x-3)^2+(23/13 x+147/26-3...

Suppose we are given two points r1 = {x1, y1} and r2 = {x2, y2} and radius R.
Suppose the center is at r = {x, y}. Then:
(x - x1)^2 + (y - y1)^2 = R^2
(x - x2)^2 + (y - y2)^2 = R^2
We can parametrize the first equation:
x = x1 + R cos(t)
y = y1 + R sin(t)
We substitute that into the second equation.
(x1 - x2 + R cos(t))^2 + (y1 - y2 + R sin(t))^2 = R^2
Let x2 - x1 = Δx, y2 - y1 = Δy. Then:
(R cos(t) - Δx)^2 + (R sin(t) - Δy)^2 = R^2
R^2 - 2R(Δx cos(t) + Δy sin(t)) + Δx^2 + Δy^2 = R^2
Δx cos(t) + Δy sin(t) = (Δx^2 + Δy^2)/(2R)
Let z = Δx + iΔy. Then:
|z|cos(t - arg(z)) = |z|^2/(2R)
cos(t - arg(z)) = |z|/(2R)
t - arg(z) = ±arccos(|z|/(2R)) + 2πn, n ∈ ℤ
t = arg(z) ± arccos(|z|/(2R)) + 2πn, n ∈ ℤ
We are only interested in two values. So:
t = arg(z) ± arccos(|z|/(2R))
Now we look at both cases. Before that, though, let's recap the trigonometric functions of those terms.
cos(arg(z)) = Δx/√(Δx^2 + Δy^2)
sin(arg(z)) = Δy/√(Δx^2 + Δy^2)
cos(arccos(|z|/(2R))) = √(Δx^2 + Δy^2)/(2R)
sin(arccos(|z|/(2R))) = √(1 - (Δx^2 + Δy^2)/(2R)^2)
Then:
cos(t) = Δx/(2R) ∓ Δy√(1/(Δx^2 + Δy^2) - 1/(2R)^2)
sin(t) = Δy/(2R) ± Δx√(1/(Δx^2 + Δy^2) - 1/(2R)^2)
So:
x = x1 + R cos(t) = x1 + Δx/2 ∓ (Δy/2)√(4R^2/(Δx^2 + Δy^2) - 1)
y = y1 + R cos(t) = y1 + Δy/2 ± (Δx/2)√(4R^2/(Δx^2 + Δy^2) - 1)
Now, let's see if we can notice anything useful if we expand Δx and Δy.
x1 + Δx/2 = x1 + (x2 - x1)/2 = (x1 + x2)/2
y1 + Δy/2 = y1 + (y2 - y1)/2 = (y1 + y2)/2
Δx^2 + Δy^2 = (x2 - x1)^2 + (y2 - y1)^2 = x1^2 + y1^2 + x2^2 + y2^2 - 2(x1x2 + y1y2) = |r1|^2 + |r2|^2 - 2r1·r2 = |r2 - r1|^2
Thus, we get:
x = (x1 + x2)/2 ∓ ((y2 - y1)/2)√(4R^2/|r2 - r1|^2 - 1)
y = (y1 + y2)/2 ± ((x2 - x1)/2)√(4R^2/|r2 - r1|^2 - 1)
Or, even more simply:
r = (1/2)(r1 + r2) ± (1/2)√(4R^2/|r2 - r1|^2 - 1)R(r2 - r1)
Here R(r2 - r1) = {0, 0, 1}⨯(r2 - r1) is the vector (r2 - r1) rotated by 90° in the counterclockwise direction.
I've used this for your case, and it produces the correct results, as you can see.

image.png
#

Quite an interesting problem!

#

The exact solutions in your case are r = {29/2 ∓ (13/2)√(901/349), 63/2 ± (23/2)√(901/349)}.

willow forge
willow forge
#

And you know the two congruent sides and can calculate the base with the distance formula, from which you can calculate the height.

#

The height of triangle AOB passes through the midpoint of AB and is normal to it, which is enough to determine its equation, and then the actual length of the height will tell you where the center is. Or two possible locations, from which the diagram will tell you which one is the center by telling you whether the center is above or below AB.

charred badge
#

I don't have the angle

willow forge
#

If you know the lengths of all three sides of a triangle, you know everything about the triangle. Three side lengths completely determine a triangle.

charred badge
willow forge
onyx torrent
#

Oh, are you proposing a geometric construction?

#

Yeah, solving this with geometry is quite easy. Even with a circle and straight edge - literally just draw 2 circles.

charred badge
#

So your way seems the only way

#

But seems like a very, well not simple solution, compared to my level of maths we're in, so I think I can just use wordmat and say what it did instead of the 100 steps

onyx torrent
#

It's just that if you want the exact values of coordinates instead of an abstract point, the algebraic approach is better.

willow forge
#

I mean, my approach can also yield exact coordinates.

charred badge
#

I can plot it into geogebra yeah but I don't think my teacher will appreciate that

onyx torrent
willow forge
#

Look, given a radius and two points, there are always exactly zero, exactly one, or exactly two circles of that radius passing through those points, for 2r < dist(A, B), 2r = dist(A, B), and 2r > dist(A, B), respectively.

onyx torrent
#

Yup.

charred badge
#

I have math in a different language so the barrier in terms of math lol have to read what you're saying 3 times to get it

willow forge
charred badge
#

diameter

willow forge
# charred badge 2r

Right. So if the distance between two points is more than the diameter of the circle, then the circle can't possibly contain both points, right?

charred badge
willow forge
# charred badge yh?

And if the distance between two points is exactly the diameter, then the only circle which can contain both points is the one with the center at the midpoint of the segment defined by those points, right?

#

That is.

charred badge
#

Yes

willow forge
#

Oh good, I was worried "segment" or "midpoint" might throw you off.

#

And if the diameter of a circle is greater than the distance between two points, then there are two possible circles that can contain them. The explanation for this one involves those isosceles triangles.

charred badge
#

no im just confused how knowing all the sides will get me x and y of the circles midpoint

willow forge
charred badge
#

yes

#

2 cathetes being 25

willow forge
#

Which means that its height is the line which bisects (I can explain this word) the base and is at a right angle to it.

#

The perpendicular bisector is what such a line is called.

#

Like, in general if you don't know what a word means, you can ask me to explain the word.

charred badge
#

perpendicular bisector?

willow forge
#

Right. Okay. So "perpendicular" means "at a right angle to", and "bisector" means "splits in half/intersects (passes through) the midpoint".

charred badge
#

ok

#

so what can i use this for

#

i know we have 91 40,5 and 40,5 degress

willow forge
#

We actually don't need the angles, per se.

charred badge
#

ok

willow forge
#

Just the sides.

#

Okay, here's where we're going; if you split an isosceles triangle in half, you get two congruent right triangles.

#

And the height of those right triangles equals the height of the isosceles triangle.

charred badge
#

25, 25 & 39,83717

Are the sides

So you say split in half so we have 25 25 and 19.918585

willow forge
#

Well, no.

#

We have 25, whatever the exact value of that decimal thingy is, and the height which we have yet to calculate.

#

And actually that distance is inaccurate.

charred badge
#

wut how

willow forge
#

Show your work.

charred badge
#

√((43-3)^2+(25-38) )≈39,83717

#

oh yh forgot squared the last

willow forge
#

Where the hell is 43 coming from?

charred badge
#

2sec

#

√((26-3)^2+(25-38)^2 )≈26,41969

#

there we go

willow forge
#

We can keep it in exact form as sqrt(698).

#

So, we know from the Pythagorean theorem that (sqrt(698)/2)^2 + h^2 = 25^2, right?

charred badge
#

but isnt that only when we have 90 degree corner

#

we dont here no?

#

is 91

#

or actually we dont have the degree

willow forge
charred badge
#

ahh yeah right

#

wait no we have A to B

#

not C

willow forge
#

It was a mistake.

charred badge
#

but same idea

#

yh

#

the height is then 625-√698

#

that makes no sense stare

willow forge
#

No.

#

Show your work.

charred badge
#

〖√698〗^2+x^2=25^2

no?

willow forge
#

No.

#

For one, you forgot to divide by two, but that doesn't fix the problem.

charred badge
#

oh yh divide by 2

willow forge
#

So after you fix that, what's the next step?

charred badge
#

so 21,22499 is height

#

what do i do with that stare

willow forge
charred badge
willow forge
#

Anyway.

#

So now we need to determine the line on which the center lies.

#

And it's the perpendicular bisector of AB, so we can determine it with point slope form. What's the coordinates of the midpoint of AB?

charred badge
#

(14,5 28,10869565217391)

willow forge
#

...how many times do I have to say "exact value"?

charred badge
#

idk how else to make it more exact

#

(14.5, -188,5+835/23)

willow forge
#

That's not the exact value, and it can't be the exact value because it's irrational.

charred badge
willow forge
#

Or, wait, is it not irrational?

charred badge
#

it just gives me straight =

#

no wobbly = or whatever u call it

willow forge
#

The midpoint coordinates are rational, why did you get that giant decimal monstrosity?

#

What's the midpoint formula?

charred badge
#

hehe ok see i didnt use that i freestyled it 💀

willow forge
#

Why would you do that? How did you do that?

charred badge
#

slope=(25-38)/(26-3)=-(13/23)
b=38-(13/23)·3=835/23

f(x)=-(13/23)x+835/23

-(13/23)·14,5+835/23=28,10869565217391

🤷‍♂️

#

made sense in the moment lol