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.
+128474 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
