Circle and a line dividing the circumference

[Navin]

Homework Statement

The line "4x + 3y -4=0 " divides the circumference of circle centred at (5,3) in ratio 1:2.whats the equation of the circle

Homework Equations

2πr
Distance of a line from point
Radius of general circle

The Attempt at a Solution

I tried finding the distance of the centre from the chord.
Then i made a relation between radius and common factor of the ratio between the two circumfernce divisions
And now I am stuck :(

 

[YoungPhysicist]

you can check out “circle functions” in detail on the internet.

 

[Navin]

you can check out “circle functions” in detail on the internet.

How do circle functions help here ?
They haven't given any angle or what's so ever .

 

[YoungPhysicist]

(Just an indication)
If you assume the central angle of the triangle made by the two points which the line and the circle and the center as alpha , the area of it will be:

Then you can use the conditions mentioned to calculate the area of this triangle,then calculate
the radius by the equation above.

 

[YoungPhysicist]

https://math.stackexchange.com/ques...umference-of-the-circle-x2y2-100-into-two-arc
Similar question

 

[Navin]

https://math.stackexchange.com/ques...umference-of-the-circle-x2y2-100-into-two-arc
Similar question

Thank you so much !

 

[SammyS]

Thank you so much !

What is your result & how did you get it?

(We always like to know after helping someone.)

 

[Navin]

Okay sorry for the delay SammyS.
So here's what i did

 

[Navin]

i used the circle function to relate the two circum ference. ( arc length = Radius × Angle in radians) and i got the value for alpha whivh was the angle i asumed.

Then i used angle sum property of a triangle and found out that twas a equilateral triangle of angle 60°(i don't think there are other types of equilaterals in a cartisian plane !)

Then i found the perpendicular distance between the centre of circle and the line which came out to be one (using distance formula between line and point)

And then i related radius and that distance by using sin 60° and got radius.

Then by using the radius formula of general circle i found C.

And hence the equation came to be
x² +y² -10x -6y +30 = 0

I pray i didnt do a calculation error

 

[SammyS]

No that's not right, and you didn't give a lot of detail on the individual steps.

i used the circle function to relate the two circumference. ( arc length = Radius × Angle in radians) and i got the value for alpha which was the angle i assumed.

I could ask what you got for the angle, but probably more important: What was the arc length you used to get the angle, and how does that arc length compare to the circumference of the circle?

Then i used angle sum property of a triangle and found out that it was a equilateral triangle of angle 60°(i don't think there are other types of equilaterals in a Cartesian plane !)

It's not clear how this triangle is oriented with respect to the circle or to the line.

Then i found the perpendicular distance between the centre of circle and the line which came out to be one (using distance formula between line and point)

What is the equation of the line which is perpendicular to the given line?

That is not the correct distance.

And then i related radius and that distance by using sin 60° and got radius.

What radius did you get?

Then by using the radius formula of general circle i found C.

What is this C?

Give this formula.

And hence the equation came to be
x² +y² -10x -6y +30 = 0

I pray i didnt do a calculation error

That is the equation of a circle centered at (5, 3), and its radius is 2.

Is that what you intended?

The graph given by @Young physicist in post (#4) is not necessarily to scale, but ifs clear that when you consider the x and y intercepts of the given line, the radius of the circle needs to be much greater than 2 to have any points in common with the given line.

 

[YoungPhysicist]

The graph given by @Young physicist in post (#4) is not necessarily to scale,

Just giving a general concept

 

[neilparker62]

Just giving a general concept

This diagram is pretty much on the mark. I would print it out and then draw the line from centre of circle to (calculated) midpoint of the chord. Mark the angles you can obtain from the circumference ratio and the rest should be simple trig!

 

[Navin]

No that's not right, and you didn't give a lot of detail on the individual steps.
I could ask what you got for the angle, but probably more important: What was the arc length you used to get the angle, and how does that arc length compare to the circumference of the circle?
It's not clear how this triangle is oriented with respect to the circle or to the line.
What is the equation of the line which is perpendicular to the given line?

That is not the correct distance.
What radius did you get?
What is this C?

Give this formula.

That is the equation of a circle centered at (5, 3), and its radius is 2.

Is that what you intended?

The graph given by @Young physicist in post (#4) is not necessarily to scale, but ifs clear that when you consider the x and y intercepts of the given line, the radius of the circle needs to be much greater than 2 to have any points in common with the given line.

Yes i just realized that my distance is wrong.it should be 5.(i forgot to root something)

Wait ill just post a detailed solutiin in a few min

 

[Navin]

Ps - is it compulsary that i type the full soln or can i just send an attachement with the handwritten working ? Someone had said it was against forum policy to do the later.just confirming

 

[Navin]

Okay so here's what i did.

 

[Navin]

So first i calculated Arc Length
Now the ratio of circumference is 1:2.
Let the common factor be x
S = r θ where θ is central angle of the minor arc
x =r ×θ...
X/θ = r (1)

Lets look at major arc

2X = r ×(2π-θ)
2X= (X/θ)(2π-θ)

2θ=2π-θ
3θ=2π
θ =2π/3

Converting to degrees we get (2 ×180)/3=120°
(Not 60 as i previously screwed up in calculations)

 

[Navin]

Now let's look at the distance tween cente and the line

This is given by | (ax + by + k)÷(root of (a² +b²|

Where a b and k are constants in the equation of line and x and y are coardinates of the point
And" | | "represent modulus

With that we get
| (4 ×5 + 3×3 -4) ÷ 5| = 5(not 1 as i previosly did)

 

[Navin]

If we look at that triangle and use angle sum prop
We get Φ +Φ +120 = 180

Φ=30

 

[Navin]

Hence now we can finaly find radius
By using sin 30 =opposite side÷hypotenuese

(1/2) =(5/r)
r =10

 

[Navin]

With this we can use the formula for radius

g² + f² +c = 10
Where -g and -h are coord nates of centre

25 +9 +c =10

c= -24

 

[Navin]

And now that we have all unknows the equation of general circle

x² +y² + 2gx +2fy +c =0

Hold on i made an error

The formula for radius is g² + f² -c = 10

Hence r = -24

 

[Navin]

Anyway so equation of circle is

x² +y² -10x -6y -24 =0

 

[Navin]

I checked the answer given and its not matching

 

[Navin]

We r suposed to get
x^2 +y^2 -10x -6y -66 =0

 

[Navin]

We r suposed to get
x^2 +y^2 -10x -6y -66 =0

Any ideas ?

 

[SammyS]

So first i calculated Arc Length
Now the ratio of circumference is 1:2.
Let the common factor be x
S = r θ where θ is central angle of the minor arc
x =r ×θ...
X/θ = r (1)

Lets look at major arc

2X = r ×(2π-θ)
2X= (X/θ)(2π-θ)

2θ=2π-θ
3θ=2π
θ =2π/3

Converting to degrees we get (2 ×180)/3=120°
(Not 60 as i previously screwed up in calculations)

That's good !

Also, the distance from (5,3) to the line 4x + 3y − 4 = 0 is indeed 5, as you found.

 

[Navin]

That's good !

Also, the distance from (5,3) to the line 4x + 3y − 4 = 0 is indeed 5, as you found.

But the equation I am getting is wrong ! Where did i make an error ?

 

[SammyS]

If we look at that triangle and use angle sum prop
We get Φ +Φ +120 = 180

Φ=30

Hence now we can finaly find radius
By using sin 30 =opposite side÷hypotenuese

(1/2) =(5/r)
r =10

That looks O.K., unless I'm missing something.

Also $\ \displaystyle \cos\left(\frac{\theta}{2}\right) = \frac{5}{R} \$ should work, for θ = 120°.

 

[SammyS]

And now that we have all unknows the equation of general circle

x² +y² + 2gx +2fy +c =0

Hold on i made an error

The formula for radius is g² + f² -c = 10

Hence r = -24

I would use the following for the formula of a circle, radius R and centered at point ( h, k ) :

$\displaystyle (x-h)^2 + (y-k)^2 = R^2$​

.
To get the equation in the form you want, expand the those squared terms and simplify.

 

[SammyS]

With this we can use the formula for radius

g² + f² +c = 10
Where -g and -h are coord nates of centre

25 +9 +c =10

c= -24

After looking at this again, shouldn't this be radius squared?
:
$(-g)^2 + (-h)^2+c = R^2 \,,\$ where g and h are coordinates of the centre and R is the radius. In your case, R = 10 .

 

[Navin]

After looking at this again, shouldn't this be radius squared?
:
$(-g)^2 + (-h)^2+c = R^2 \,,\$ where g and h are coordinates of the centre and R is the radius. In your case, R = 10 .

*face palm*
Yes...its radius square

 

[Navin]

So

*face palm*
Yes...its radius square

It will be

100 -24-9
=100-33
=66
...yep so the value of c is 66
Hence the circle will be

x^2 +y^2 -10x -6y -66 =0

Yay ! FINALY !

 

[Navin]

Thank you Sammy and A Young Physist And All those who helped me

 

[Navin]

You all get a virtual cookie
...mmmmm taste those bytes(get the pun)

 

[SammyS]

You all get a virtual cookie
...mmmmm taste those bytes(get the pun)

Let's see... How many bits are in each byte?

... and are these bits similar to crumbs - in the case of a virtual cookie ?

 

[Navin]

Let's see... How many bits are in each byte?

... and are these bits similar to crumbs - in the case of a virtual cookie ?

Speakibg of one bits in a byte...thats a good question and raises the heap of sand paradox.

Supposes i were to remove one bit from a sum of bits making a byte...would it still appear to be a byte...and if not what do we call it...then again how many bits make a byte !...very deep philosophical questions there

 

[SammyS]

I would use the following for the formula of a circle, radius R and centered at point ( h, k ) :

$\displaystyle (x-h)^2 + (y-k)^2 = R^2$​

.
To get the equation in the form you want, expand the those squared terms and simplify.

For me, this form of a circle is much easier to remember. It's basically the Pythagorean theorem.
In your case:

$\displaystyle (x-5)^2 + (y-3)^2 = 10^2$
$x^2 - 10x +25 + y^2 -6y +9 = 100$​

..

 

[Navin]

For me, this form of a circle is much easier to remember. It's basically the Pythagorean theorem.
In your case:

$\displaystyle (x-5)^2 + (y-3)^2 = 10^2$
$x^2 - 10x +25 + y^2 -6y +9 = 100$​

..

Yea but 90% of the sums in our books are based on the other form of writting it.Plus we have all those T=0 rules and that whole lot of tangents ,radial axes and blah which can be best written in the other form, hence we have to remeber the General form

Exam wise its better for us...but i like the pythagoran one more...its more...clean !

 

[LCKurtz]

Since you have completely solved this problem, and since I had a little time to mess with Maple this weekend, here's a pdf of a Maple worksheet you might find interesting:

 

[neilparker62]

All good except last - need 10^2.

 

[YoungPhysicist]

One final graphic conclusion of this post:

The circle is based on the equation discussed above,not some random numbers,which means it’s “the answer”

 

[Navin]

Since you have completely solved this problem, and since I had a little time to mess with Maple this weekend, here's a pdf of a Maple worksheet you might find interesting:

Cooooollllll !

 

[SammyS]

One final graphic conclusion of this post:

[ ATTACH=full]231144[/ATTACH]

The circle is based on the equation discussed above,not some random numbers,which means it’s “the answer”

I'm guessing you refer to post below (#40), which doesn't give a reference regarding what it is replying to.

All good except last - need 10^2.

Checking back through posts in this tread, I had to go back to post 20, or 21 to find something this may apply to.
FYI: It helps to respond to a specific post or fragment of a post by using te "Reply" feature.

Below, @Navin makes it clear that he is replying to a thread by @LCKurtz , which by the way has a link to a pdf file which includes a very accurate graph as part of a complete solution for the problem in this thread.

Since you have completely solved this problem, and since I had a little time to mess with Maple this weekend, here's a pdf of a Maple worksheet you might find interesting:

Cooooollllll !

 

[Navin]

I'm guessing you refer to post below (#40), which doesn't give a reference regarding what it is replying to.

Checking back through posts in this tread, I had to go back to post 20, or 21 to find something this may apply to.
FYI: It helps to respond to a specific post or fragment of a post by using te "Reply" feature.

Below, @Navin makes it clear that he is replying to a thread by @LCKurtz , which by the way has a link to a pdf file which includes a very accurate graphical representation of the solved problem for this thread.

Wow...this is the first time someone has analysed by words with such vigor ! +1 Navyn point to you.

You can collect +1 Navin points to get a Mega-Vyn point and subsequently go on and on to get a giga-Vyn Point.

 

[Chestermiller]

Wow...this is the first time someone has analysed by words with such vigor ! +1 Navyn point to you.

You can collect +1 Navin points to get a Mega-Vyn point and subsequently go on and on to get a giga-Vyn Point.

I did this problem completely differently. I represented the circle parametrically as
$$x=5+r\cos{\theta}$$
$$y=3+r\sin{\theta}$$
To get the intersection with the line, we can write: $$3(5+r\cos{\theta})+4(3+r\sin{\theta})=4$$or equivalently:
$$\frac{4}{5}\cos{\theta}+\frac{3}{5}\sin{\theta}=-\frac{5}{r}$$ If we let $$\cos{\phi}=-\frac{4}{5}$$and$$\sin{\phi}=-\frac{3}{5}$$we have$$\cos{(\theta-\phi)}=\cos{\psi}=\frac{5}{r}$$The values of $\psi$ that divide the circle circumference in the ratio of 2:1 are $\pi/3$ and $2\pi-\pi/3$ So, $$\frac{1}{2}=\frac{5}{r}$$or$$r=10$$

 

[neilparker62]

Checking back through posts in this tread, I had to go back to post 20, or 21 to find something this may apply to.
FYI: It helps to respond to a specific post or fragment of a post by using te "Reply" feature.
.

Noted. I think the problem was I was on page 1 of the posts and thought the last there was the last - meanwhile the discussion had moved on considerably!

 

[Navin]

I did this problem completely differently. I represented the circle parametrically as
$$x=5+r\cos{\theta}$$
$$y=3+r\sin{\theta}$$
To get the intersection with the line, we can write: $$3(5+r\cos{\theta})+4(3+r\sin{\theta})=4$$or equivalently:
$$\frac{4}{5}\cos{\theta}+\frac{3}{5}\sin{\theta}=-\frac{5}{r}$$ If we let $$\cos{\phi}=-\frac{4}{5}$$and$$\sin{\phi}=-\frac{3}{5}$$we have$$\cos{(\theta-\phi)}=\cos{\psi}=\frac{5}{r}$$The values of $\psi$ that divide the circle circumference in the ratio of 2:1 are $\pi/3$ and $2\pi-\pi/3$ So, $$\frac{1}{2}=\frac{5}{r}$$or$$r=10$$

I never thought of using parametric form...then again my trigo is as week as a staircase of sand

 

[SammyS]

Noted. I think the problem was I was on page 1 of the posts and thought the last there was the last - meanwhile the discussion had moved on considerably!

Yes.
I admit, that sort of thing has happened to me too .

 

[YoungPhysicist]

@Chestermiller Really elegant solution,but beyond my reach just now...though I can understand the concept behind this