A circle sits in the interior of the parabola with equation y = x^2/4, so that it is tangent to the parabola at two points. How much higher is the center of the circle than the points of tangency?
Let the center of the circle have coordinates $(a,b)$. Since the circle is tangent to the parabola, its radius is normal to the parabola at the point of tangency. Hence the tangent line to the parabola at $(a,b)$ is $y - b = \tfrac{1}{2}a(x-a)$. Since this line is tangent to the circle as well, we have $x^2 + (y-b)^2 = (x-a)^2 + \tfrac{1}{4}a^2$, which simplifies to $x^2+y^2 - 2ax - 4b = 0$. Therefore $a = \tfrac{1}{2}x$, so we have $y-b = \tfrac{1}{4}x^2$, and $x^2 + y^2 - x^2 - 8b = 0$, which means that $b = \tfrac{1}{8}x^2$. Therefore the center of the circle is at $y = \tfrac{1}{8}x^2 + \tfrac{1}{8}a^2$, which is $\boxed{\frac{1}{16}a^2}$ higher than the point of tangency.
Valerie
There is of course an infinite number of such circles that one can construct
I tried 3 different cases, with point on the parabola of (2,4), (4,4) and (8,16)
in each case I found the slope of the tangent at the point, which led to
the slope of the normal and the equation of the normal,
the centre of the circle will be the y-intercept of the normal
In each case I found the difference between the height of the centre
and the point of tangency to be 2
Working on the general case, let me know if you need it.
of course, let's ignore the gibberish by the AI.