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# Parabola Circle Problem

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A circle passes through the point \((0,1),\) and is tangent to the parabola \(y=x^2\) at \((2,4).\) Find the center of the circle.

Any help would be greatly appreciated. Thank you!

Apr 3, 2020

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The  points  (2,4)  and ( 0,1)   are  on  the circle

The slope of the tangent line to the parabola  at (2,4)   =  4

A line that  perpendicular to  this  tangent line will have a slope of  (-1/4)

And the equation  of this line will  be  y = (-1/4)(x - 2) + 4  ⇒  (-1/4)x + 1/2 + 4  = (-1/4)x + 4.5  =

-x/4 + 4.5

And this line will pass through the  center of the  circle

Let  m  be   the   x coordinate of the center

Then  y = -m/4 + 4.5      will  be  the  y  coordinate of  the  center

And  this point will be equidistant from  ( 0.1)  and (2,4)

So  using the distance  formula  we can  set  these   distances equal and solve  for  m

(m - 0)^2  +  (-m/4 + 4.5 - 1)^2  =  (m -2)^2  +  ( -m/4 + 4.5 - 4)^2   simplify

m^2  +  ( 3.5 - m/4)^2  = m^2  - 4m + 4  +  (.5  - m/4)^2

12.25 - (7/4)m + m^2/16  =  -4m + 4  +  .25 - m/4 + m^2/16

12.25  - (7/4)m  =  -4m + 4.25 -m/4

12.25 - 4.25 = (7/4 - 4 - 1/4) m

8  = -(5/2)m

8(-2/5)  =  m   = -16/5  =  -3.2   =  x  coordinate of the  center

And  the y coordinate of the  center =  -(1/4)(-16/5) + 4.5  = (16/20) + 4.5  = (4/5)  + 4.5  = .8 + 4.5  = 5.3

So......the  center  =   (-3.2, 5.3)

The  radius^2 =  (3.2)^2  + (5.3 -1)^2  = 28.73

The  equation of the  circle  is

(x + 3.2)^2  + ( y - 5/3)^2  = 28.73

Here's  a graph  :  https://www.desmos.com/calculator/4zun6t2nwj

Apr 3, 2020