# Circle Problem

• Mar 16th 2010, 07:08 PM
Solaris123
Circle Problem
Im currently studying for an exam which frankly is way over my head, so I want to see who can help me, right now with this particular problem. I need not only the solution but how to solve and how to apply it to other situations. Thank you and have a good day.

Denoted by D, the domain {(x,y)}| x>=0, y>=0}

Assume a circle C contained in D touches parabola y=(1/2)x^2 at the point (2,2) and also touches the x-axis. Find the radius of C.

Please show all steps and explain them to me as I might need to apply this to other problems.
• Mar 16th 2010, 08:05 PM
Solaris123
Just a quick note, I will not have a calculator or any reference available so rather than the answer I must have the method you use to get to the answer, even if you dont solve it but can tell me how to solve it then I would be a happy man because right now Im failing in getting any real answers.
• Mar 16th 2010, 08:40 PM
sa-ri-ga-ma
i)By differentiating the equation of parabola, find the slope m at the point P.(2,2).
ii) Find the equation tangent at (2,2) using the formula (y-y1) = m(x-x1)
iii) Find the point of intersection of the tangent and the x-axis. (x,0)
iv) Find the distance between the point of intersection X and P(2,2)
v) From the slope find the angle between the tangent and the x-axis.
vi) If C is the center of the circle, then CPX is a right angled triangle.
vii) The angle PXC is half the angle made by the tangent with x-axis.
viii) tan(PXC) gives you the required result.
• Mar 16th 2010, 09:11 PM
Solaris123
Just making sure, this would work with any circle that touches only one point of a parabola or it just work in this case because it touches the x-axis?
PS: I have tried you method and all seems to be going well till step v). If you could show me with this problem how it done it would be great.
• Mar 16th 2010, 09:24 PM
sa-ri-ga-ma
Quote:

Originally Posted by Solaris123
Just making sure, this would work with any circle that touches only one point of a parabola or it just work in this case because it touches the x-axis?
PS: I have tried you method and all seems to be going well till step v). If you could show me with this problem how it done it would be great.

This method works only when the circle touches the x-axis.
Slope of the tangent is 2. So tanθ = 2
The angle made by the tangent with x-axis is 63.4 degrees. Half of that angle = angle PXC = 31.7 degrees.
PC/PX = tan(31.7).
So the radius PC = PX*tan(31.7)
• Mar 17th 2010, 03:31 AM
Solaris123
I wanted to ask to see if anyone else had a method that didnt involve tangents? I will lack a calculator and finding those angles to do this seems a little unwidely seeing how this arent special right triangles, thus if someone else has another method which you dont need a calculator to solve it would be great.
• Mar 17th 2010, 03:11 PM
Solaris123
Teacher confirmed it
My teacher confirmed I had to find the solution by another method other than this because he wont let us use calculators. Is there anyway to use the definition the formula of a circle to solve this?
• Mar 17th 2010, 05:35 PM
skeeter
I know you have a sketch ...

circle center is $\displaystyle (r,y)$

slope from $\displaystyle (r,y)$ to $\displaystyle (2,2)$ has magnitude $\displaystyle |m| = 1/2$

let $\displaystyle a$ = vertical distance from $\displaystyle (2,2)$ to $\displaystyle (r,y)$

$\displaystyle 2+a = y$

then $\displaystyle 2a$ = horizontal distance from $\displaystyle (r,y)$ to $\displaystyle (2,2)$

$\displaystyle r + 2a = 2$

also, using Pythagoras ... $\displaystyle r = \sqrt{a^2 + (2a)^2}$

$\displaystyle r = \sqrt{5} \cdot a$

$\displaystyle \sqrt{5} \cdot a + 2a = 2$

$\displaystyle a(\sqrt{5} + 2) = 2$

$\displaystyle a = \frac{2}{\sqrt{5} + 2}$

$\displaystyle r = \sqrt{5} \cdot \frac{2}{\sqrt{5} + 2} = \frac{2\sqrt{5}}{\sqrt{5} + 2}$