Circle inscribed in a parabola

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SUMMARY

The discussion focuses on finding the center of a circle with a radius of 1 inscribed in the parabola defined by the equation y=x^2. The center of the circle is determined to be at the coordinates (0, 5/4). The participants explore the relationship between the equations of the circle, (x-h)^2 + (y-k)^2 = 1, and the parabola, y=x^2, to derive the necessary values for k. The solution involves differentiating the equations and solving for the intersection points, ultimately leading to the conclusion that k can be found by ensuring the derivatives of both functions match at the points of tangency.

PREREQUISITES
  • Understanding of parabolic equations, specifically y=x^2
  • Familiarity with circle equations in the form (x-h)^2 + (y-k)^2 = r^2
  • Basic calculus concepts, including differentiation
  • Knowledge of solving systems of equations
NEXT STEPS
  • Study the method of finding points of tangency between curves
  • Learn about implicit differentiation and its applications
  • Explore the geometric properties of circles and parabolas
  • Practice solving systems of equations involving quadratic functions
USEFUL FOR

Students studying calculus, particularly those interested in geometry and optimization problems involving curves, as well as educators looking for examples of inscribed shapes in conic sections.

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Homework Statement


A circle with radius 1 inscribed in the parabola y=x^2. Find the center of the circle.

Homework Equations


equation of the parabola: y=x^2
circle: (x-h)^2 + (y-k)^2 = 1

The Attempt at a Solution


After ghosting the forums and reading through every post with this exact same question, I can't quite get to the answer, which is (0,5/4).

i've assigned notation to the coordinates of the center of the circle C(h,k), and the the two points where it intersects the parabola P(x1,(f(x1)), and Q(-x1,f(-x1)). where f(x) functions y: y=f(x)=x^2.

since the center of the circle lies on the y-axis, therefore h=0. thus, the new equation of the circle is:
x^2 + (y-k)^2 = 1

by relating the parabola to the equation of the circle, i get:
x^2 + (x^2-k)^2=1

this is where i am a bit confused about how to proceed to find the value of k. i just finished calculus 1 (differentiation) this spring quarter and can't for the life of me think of how to solve this problem which is quite infuriating. from reading other posts here, I've managed to arrive at what various responses to this same question have instructed me to do, but am still lost as how finish it... here's what i got.

from differentiating the equation of the circle, i get:
2x + 2(x^2-k)(2x-0)=0
-> 2x + 4x(x^2-k)=0
-> 4x(x^2-k)=-2x
-> 1 = -(1/2(x^2-k))
-> 2(x^2-k) = -1

i've also tried expanding the equation before differentiating, i get
x^2 + x^4 -2x^2k + k^2 = 1

from grouping like-terms and factoring x^2, i get:
x^4 + (1-2k)x^2 + k^2 = 1

and here i am again am stumped as how to proceed... I've tried using the quadratic equation where a=1, b=(1-2k) (?) and c=1 but it that doesn't seem right either.
 
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I love problems like this. :smile:

I can't get any farther with the problem with what you've done, so I've started of in a different way.
So what you want to find is k since you know that h = 0. But in order to find k, you also need to find some x and y for your circle equation (y - k)2 + x2 = 1 so you can solve for k. I wouldn't substitute the equations you have and I don't think that will help you at all. Instead, solve for y and get a function of x, say g(x), for the circle (make sure you get the right function that describes the kind of circle, or part of circle actually, you have) and let the parabola be f(x) = x2 as you already did.

The derivatives of f(x) and g(x) will be the same at the two points where the circle touches the parabola, so solving the two equations together will give you an x and y like the (x1,(f(x1)) you mentioned. Substitute them in (y - k)2 + x2 = 1, or better yet your new function g(x) for the circle, and you will get k.
 
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