Finding the radius of a Circle Touching a Parabola and the Axis

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May 26th 2011, 03:08 PM
Zellator

Finding the radius of a Circle Touching a Parabola and the Axis

Hi Forum!

I have a little problem with a question:

What is the radius of a a circle that touches the parabola 1/2x² at (2,2) and the x-axis?

Here the derivative of our parabola is y=x
And if we get
$(x-2)^2+(y-2)^2=(x-v)^2+y^2$

This could give us the common point between those two circles; that would be the center of the circle that we want.
But this gives me a strange equation, that I really don't know what it is for.

Or,
Get the center (a,b)
and use it at the circle equation.
$(x-a)^2+(y-b)^2=r^2$

But where does (2,2) enters here?
I saw something about using both at the same equation, like:

$(a-2)^2+(b-2)^2=r^2$
Is this possible? Aren't x and y fixed values here?

Thanks! I can't find a good source to research on this.
May 27th 2011, 04:09 AM
emakarov

Quote:

Originally Posted by Zellator

What is the radius of a a circle that touches the parabola 1/2x² at (2,2) and the x-axis?

Here the derivative of our parabola is 2
And if we get
$(x-2)^2+(y-2)^2=(x-v)^2+y^2$ What is v?

This could give us the common point between those two circles; that would be the center of the circle that we want. Which two circles?
But this gives me a strange equation, that I really don't know what it is for. Why is it strange?

It's not easy to understand your thought.

Hope this helps.

https://lh3.googleusercontent.com/-l.../parabola2.png
May 27th 2011, 04:35 AM
Zellator

Oh, sorry to not make myself clear, emakarov.

v is the x coordinate of the place where the circle touches the x-axis. The Xo in your graph.
I read somewhere that you could get get center of the circle by using two circles in the places where we have points, the parabola and the x-axis.
Using this formula I'll get a equation with some unwanted X's and Y's and V's, so it's probably not the best idea to use it.

I really thought about using a perpendicular here. But what is the easiest way to get it?

I would use it like f(x-2)+2 using the translation principles.
$y'=-(1/2)x -2$

So

$y'=-(1/2)(x-2)+2=-(x/2)+3$

But what about the tangent line?

To get it you used $y-2=2(x-2)$ right?
I was kind of confused in that terms too, since I hadn't seen calculus in some time.

Ok, I will give it a go one more time.
Thanks again, emakarov!
May 27th 2011, 04:45 AM
emakarov

You have two equations on x0, y0: the first says that the distance from (x0,y0) to (2,2) equals the distance from (x0,y0) to (x0,0), and the second says that (x0,y0) lies on the graph of y = 3 - x/2.
May 27th 2011, 02:36 PM
Zellator

Quote:

Originally Posted by emakarov

You have two equations on x0, y0: the first says that the distance from (x0,y0) to (2,2) equals the distance from (x0,y0) to (x0,0), and the second says that (x0,y0) lies on the graph of y = 3 - x/2.

Hi emakarov!
I've managed to do it!!!
Yes, it was quite easy with your explanation! Thank you so much!
There's another way that I don't understand well https://lh6.googleusercontent.com/-E...rx6Zm4/YES.jpg

What's that -1 and that 2?
I'm not sure, but the b^2-5b+5 could be acquired in some way, replacing a for b in the first equation.
This way it was answered in terms of y, it seems easier to use x though!

Thanks emakarov!