Calculating the Equation of a Circle Tangent to x=sqrt(y) at (10,2)

[say_physics04]

equation of a circle...

a circle passes to points (2,0) and (18,0) and tangent to the graph of x=squareroot of y(half parabola)... the answer is supposed to be C(10,2)


can you help me in finding a way to solve this?? please help me to know the exact procedure on doing this...

thanks...

 

[cristo]

Well, you've not asked a question!

 

[VietDao29]

a circle passes to points (2,0) and (18,0) and tangent to the graph of x=squareroot of y(half parabola)... the answer is supposed to be C(10,2)


can you help me in finding a way to solve this?? please help me to know the exact procedure on doing this...

thanks...

No, we don't solve problem for people here. We only guide them to do it. We are not paid to give students full solution, or exact procedure (this is just for their own sake ).

You should show us that you've tried it, even if it's wrong (well, probably, if you get it correct, I don't think you will ask). So that we can know that you did put some efforts in solving the problem. It's just not fair if we do the homework for you, right?

So, have you tried anything?

Well, you've not asked a question!

It's the title.

 

[HallsofIvy]

But he said "the answer is supposed to be C(10,2)". How is C(10,2) the equation of a circle?


say physic04, you know that the center is on the line x= 10 (do you see that?). You know that one point on the circle is also on y= x² with x> 0 and the circle and parabola have the same tangent line there. What is the derivative of y= x²

 

[symbolipoint]

HallsOfIvy has a good method described. On the contrary, notice that this question is in the PreCalculus board and therefore derivatives might be assumed not appropriate for the question. Also, attempting a Limits analysis might also be unfitting. Still, if you understand the beginning-calculus concept of the derivative, his method is perfect. This makes us wonder if the introductory study of Limits is commonly included in PreCalculus courses these days. (By "PreCalculus", I mean that course which includes a combination of Trigonometry and the level of Algebra just a step above "Intermediate Algebra").

 

[HallsofIvy]

The tangent to a circle can be handled without calculus but I'm not so sure about the tangent to a parabola. Pascal had a method which was close to finding the derivative but, as I recall, it was extremely difficult.

 

[say_physics04]

ok,, the (10,2), the center, is given as a hint by my prof,, for the purpose of checking our answers...

we are asked to find the equation of the circle,, in general or center-radius form..

i've tried solving it,, i represented my C as (x,y),, i used systems of equations and i got x=10,, but next,, my y variable is always cancels out in the equations,, so i can't show that it must be 2...

can u guide me now,, in solving this and show that the C is (10,2)?

anyways,, for added info,, how to use that derivative method?? jst curious..

thanks to all,, i really need this tom...

 

[say_physics04]

ok,, thanks to all...

the C is (10,2) given by my prof for the purpose of self-checking our answers..

we are asked to find the solution in getting the equation of the circle..

ive tried solving it by letting C(x,y) and using systems of equations and i got x = 10 but my y variable always cancel out...

can u guide me now?? pls,, i need this asap...

thanks to all,,

anyways,, I'm curious at the derivative method,, can u teach me that also??

 

[cristo]

ok,, thanks to all...

the C is (10,2) given by my prof for the purpose of self-checking our answers..

What does C(10,2) mean?

 

[Gib Z]

The tangent to a circle can be handled without calculus but I'm not so sure about the tangent to a parabola. Pascal had a method which was close to finding the derivative but, as I recall, it was extremely difficult.

I know that method! mathwonk taught it to me on the forums before :) but he said it only works for polynomials, which a circle is not. But it's still good :)

 

[HallsofIvy]

Cristo, C(10,2) is the circle with center at (10,2) and radius \sqrt{68} (since it must pass through (2,0) and (18,0)).

 

[say_physics04]

Cristo, C(10,2) is the circle with center at (10,2) and radius \sqrt{68} (since it must pass through (2,0) and (18,0)).

yeh that's right...

first of all,, thanks to the replies that "somehow" help me

and,, i hope that no arguing of nonrelated things will be done in my thread...

anyways,, i already solved the problem in my own,,, jst few distance formulas

dont question the homework helpers,, let them do what they want,, and me too,, doesn't agree in solving others' problems completely...

anyways,,, thanks to all...