+0  
 
0
63
1
avatar

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
 #1
avatar+111321 
+1

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

 

 

 

cool cool cool

 Apr 3, 2020

15 Online Users

avatar
avatar