Find Center of Circle in Parabola y=x^2

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SUMMARY

The discussion focuses on finding the center of a unit circle inscribed in the parabola defined by the equation y=x^2. The key approach involves determining the intersection points of the circle's equation, (y-h)^2+x^2=1, with the line y=0. The solutions indicate that for |h|<1, the circle intersects the x-axis at two points, while for |h|=1, it is tangent to the x-axis. This method can be extended to find the specific center of the circle inscribed within the parabola.

PREREQUISITES
  • Understanding of parabolic equations, specifically y=x^2
  • Knowledge of circle equations in Cartesian coordinates
  • Familiarity with solving simultaneous equations
  • Basic concepts of geometry related to tangents and intersections
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  • Study the properties of parabolas and their geometric implications
  • Learn about the intersection of curves in analytic geometry
  • Explore the concept of tangents to circles and their applications
  • Investigate the use of graphical methods to visualize circle and parabola intersections
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I'll do a much simpler example and see if it helps you. Imagine you want to find the point where a unit circle centered at some point on the y-axis is tangent to the x-axis. Clearly we want the center at (0,1) or (0,-1). But another way to find this is to look at where a unit circle centered at (0,h) intersects the line y=0. This is found as the set of simultaneous solutions to:

(y-h)^2+x^2=1
y=0

which is just the set of solutions to:

x^2+h^2=0

If |h|>1, there are no solutions (which reflects the fact that the circle doesn't touch the x-axis), if |h|<1, there are two solutions (the circle straddles rhe x-axis and intersects it in two points), and if h=1 or h=-1, there is one solution, where the circle is tangent, which is just what we expected. Can you figure out how to extend this to your question?
 
errr..im still confused, although I do understand what youve done I don't know how to apply it.
 

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