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Points O and P on the graph of x^2 = 20y - 100 are the centers of circles tangent to both axes.  These circles intersect at (0,k).  Find k.

 Dec 18, 2019
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x^2  =  20y - 100

 

We need to  find the points  on this parabola   (a, a)     and   ( -a, a)

 

So

 

a^2  = 20a - 100

a^2 - 20a +100 = 0

(a - 10)^2  = 0

a = 10

So....the center of one circle is  (10, 10)

 

And by symmetry, the center of the other circle will be  (-10, 10)

 

The radii of both circle will also = 10....so....we have  these equations

 

(x - 10)^2  +  ( y - 10)^2  =  100

(x + 10)^2 +   ( y -10)^2  = 100       set these equal  and we have that

 

(x - 10)^2  + ( y - 10)^2  = (x + 10)^2  + (y - 10)^2    subtract (y -10)^2 from both sides

 

(x - 10)^2 =  (x + 10)^2

 

x^2 - 20x + 100 =  x^2 + 20x + 100

 

-20x  = 20x

 

x = 0.....this is the x intersection of the circes

And the y intersection  can be found as

(0- 10)^2  + (y -10)^2  = 100

100  + (y -10)^2  =100

(y - 10)^2  = 0

y =10

 

So.....the intersection  point is  (0,10)

 

Here's a graph :  https://latex.artofproblemsolving.com/6/6/c/66cb67543703cafc48a84fd49998af9ddbc7d9c1.png

 

 

cool cool cool

 Dec 18, 2019

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