Circle and a line dividing the circumference

[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