Find radius, center of circle passing through (2,0) and....

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111111

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Can someone post detailed steps/the answer to the following problem:

Find the radius and center of the circle that passes through the points (2,0) and (18,0) and is tangent to the curve x = square root of y

Thanks a lot in advance, I need this ASAP
 
Re: HELP

111111 said:
Can someone post detailed steps/the answer to the following problem:

Find the radius and center of the circle that passes through the points (2,0) and (18,0) and is tangent to the curve x = square root of y

Thanks a lot in advance, I need this ASAP
Click to expand...
I suggest very strongly that you first demonstrate your efforts. So far as I know from what you have given us, you want us to do your homework. That's not really the plan.

Circles look like this:

(x-h)^2 + (y-k)^2 = r^2

You are given a horizontal chord of the circle. This is a great hint. You can spot the x-coordinate of the circle's center by just looking at those two points. You should not let such glorious information go to waste.

Note: Do you remember from your first geometry class that a perpendicular bisector of any chord passes through the center of the circle? It's an old carpenter's trick to find the center of a given circle.
 
I appreciate your help, but I know a lot about circles and have done coutnless problems on them.

Can you please just give me a hint?

All I need is the Y coordinate point of the center and I can find the rest.

I realize that as the midpoint of 2,0 18,0 is 10,0 and that the circle's coordinates are 10, k

Any hints on how to find k?

PS don't categorize me as someone who wants someone else to do their homework, I do a lot and am just asking for a small bit of help. Sorry, I should have specified earlier.
 
I am not categorizing you as anything. I made no value judgments. Please don't assume any.

The fact of the matter is, you have shown no work at all, excepting that which has been suggested by me. What have you done on your own?

All you need is the center of the circle? That's right. That's the interesting part of the problem. If I tell you that, you won't learn anything.

So, what's your plan? How do you plan to go about finding the center of the circle.

Hint: The y-coordinate of the center is less than 100. Why is that? What good does it do us?
 
This is really one of the most interesting problems posted lately.
Yes indeed, the value of k is a good bit (a great bit) less than 100!
Look at this graphic.
tanparahi5.gif

I had to use a CAS to find that the center is (10,2).
The algebra was just too much for me. That is to say, I know the mathematics that generates the solution, but the algebra defeats me. I would welcome a simple approach.
 
We know from the given data that the x-coordinate of the center is 10. We need to find the y-coordinate.

I used:

\(\displaystyle \L\\(\sqrt{y}-10)^{2}+k^{2}=64+k^{2}\)

These lead to y=4 and x=2, the point where the circle is tangent with the parabola.

Knowing this:

\(\displaystyle \L\\64+(4-k)^{2}=64+k^{2}\Rightarrow{16-8k}\)

k=2

The center of the circle is then (10,2)

Using this we find the radius is \(\displaystyle \L\\2\sqrt{17}\)

The circle equation is:

\(\displaystyle \L\\(y-10)^{2}+(x-2)^{2}=68\)


circlend3.jpg
 
galactus said:
We know from the given data that the x-coordinate of the center is 10. We need to find the y-coordinate.

I used:

\(\displaystyle \L\\(\sqrt{y}-10)^{2}+k^{2}=64+k^{2}\)
Click to expand...

Thank you I appreciate it, but I'm not sure where you got this formula.

I understand the first term but not where the final three come

Why did you add k squared to the right squared? Where does 64 come from? Can you further explain?
 
We're told \(\displaystyle \L\\x=\sqrt{y}\)

In the circle formula, we already know h=10.

\(\displaystyle (\sqrt{y}-10)^{2}+(0-k)^{2}=r^{2}\)

But, from the given data, \(\displaystyle r^{2}=(18-10)^{2}+(0-k)^{2}=64+k^{2}\)
 
Can you explain what you mean by "it leads to x =2 and y = 4"?

Still confused
 
You realize, don't you, that you still haven't actually shown any work of your own?

Substitute known values.
Work the algebra.

Show us what you get.
 
111111 said:
I appreciate your help, but I know a lot about circles and have done coutnless problems on them. Can you please just give me a hint?
Click to expand...
You say you "know a lot about circles", but don't recognize the standard circle equations...? You are given many hints, but then ask the tutors to "just give me a hint"...?

Most (legitimate) tutors try to work with the student, helping the student learn how to stand (and compute) on his own. Please help us help you by showing some effort of your own.

Thank you for your consideration.

Eliz.
 
You do realize that I haven't had any Calc? This problem was in a Finite Math class

I solved for r squared in 2 different ways and found the value for h, but the problem is that we can't solve for k
 
111111 said:
I solved for r squared in 2 different ways and found the value for h, but the problem is that we can't solve for k
Click to expand...
I actually agree with you on this one. I see no way that galactus can explain the last equation. I wish he would.
It seems to me as if we must use \(\displaystyle \left( {\sqrt y - 10} \right)^2 + \left( {y - k} \right)^2 = 64 + k^2\) is the correct equation. That is the way I did it, using the fact that each ‘normal’ to the curve has slope \(\displaystyle \frac{{ - 1}}{{2x}}.\) I found a system of two equations is y & k. They were polynomials of degrees 4 & 3. Using a CAS I found the solution to be y=4 & k=2. But I see no simple way around this. I see no way for this to be a simple “finite mathematics” problem!
 
There you have it. See how much better it goes when you give us something to work with?
 
I see no way that galactus can explain the last equation
Click to expand...

You are so right, pka, I can't truly justify it. The mere fact it worked is a

fluke, I suppose.

I looked back on the paper I was using to solve this problem and I do

have \(\displaystyle (\sqrt{y}-10)^{2}+(y-k)^{2}=64+k^{2}\). For some

reason, when I solved it I used 0 instead of y and it

happened to come out to 4. :oops: Go figure. It coincidentally came out

to the correct answer. That's what you call falling in s**t and coming out

smelling like a rose. You said you would like to see a simpler solution,

pka. Well, there it is. You didn't say anything about adhering to any

mathematical rules :lol: :wink: Haha, laugh, kid.

I have decided to liberalize math. Too many absolutes. :D

Oh well, it was a fun problem. Hey 111111, when you turn this in, how

about letting us know what your teacher came up with and what he/she

has to say about it.
 
we didn't really get an explanation without using Calc

We got credit for what we had done
 
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