A circle rests in the interior of the parabola with an equation y=x^2 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?
Thank you very much!
Is there any more info given for this question.......like the size or equation for the circle? I think there are MANY circles that would fit the bill.....do you need an answer in terms of the circle size/equation?
Because of symmetry, the center of the circle will lie at (0, k)
Let the equation of the circle be
x^2 + (y - k)^2 = r^2
x^2 + y^2 - 2ky + k^2 = r^2
Using implicit differentiation, the slope of any tangent line to the circle is given by
2x + 2yy' - 2ky' = 0
x + yy' - ky' = 0
y' ( y - k) = -x
y' = [ x ] / [k - y]
The slope of the tangent line to the parabola at any point = d/dx (x^2) = 2x
So....equating these slopes, we have that
2x = x / [ k - y]
2 = 1 / [ k - y]
k - y = 1/2
y = k -1/2 this is the y coordinate of the intersection of the parabola and the circle
So.....the difference in the height of the circle's center and the tangent point(s) is just the difference in their y values
k - (k - 1/2) = 1/2 unit