radius of circle tangent to the parabola y=x^2

Question

In the following diagram, what's the amount of r=?

I supposed the center of the small circle to be $(0,y)$. So I have its distance from $(x, x^2)$ is $(x-0)^2+(y-x^2)^2=1400^2$ this question must have just one answer so the discriminant of the equation must be zero. But then I don't have any idea. how can I relate this to the bigger circle :(

Answer 1

You have the lower circle of radius $1400$. Assuming its center is $(0, a)$ then its equation is

$ x^2 + (y - a)^2 = 1400^2 $

At the tangency points $y = x^2 $, therefore

$ x^2 + (x^2 - a)^2 = 1400^2 $

Expanding, this becomes,

$ x^4 + x^2 (1 - 2 a) + a^2 - 1400^2 = 0 $

Since we want only one intersection point (actually two, but a single value of $x^2$), the discriminant must be zero, hence

$ (1 - 2 a)^2 - 4 (a^2 - 1400^2 ) = 0 $

Simplifying this becomes

$ 1 - 4 a + 4 (1400^2 ) = 0 $

From which,

$ a = \dfrac{ 1 + 4 (1400^2) }{4} = 1400^2 + \dfrac{1}{4} $

Now, we can move to the upper circle, its center is at $( 0, a + r )$ and radius $r$, thus its equation is

$ x^2 + ( y - (a + r) )^2 = r^2$

Again at the points of tangency, $y = x^2 $, so

$ x^2 + (x^2 - (a + r) )^2 = r^2 $

which simplifies to

$ x^4 + x^2 (1 - 2 (a + r) ) + (a + r)^2 - r^2 = 0 $

And we want the value of $x^2$ that solves this quadratic to be a single value, so again, the discriminant must be zero,

$ ( 1 - 2(a + r) )^2 - 4 ( a^2 + 2 a r ) = 0 $

Expanding,

$ 1 - 4 (a + r) + 4 (a^2 + 2 a r + r^2 ) - 4 (a^2 + 2 a r ) = 0 $

and this simplifies to

$ 1 - 4 (a + r) + 4 r^2 = 0 $

Knowing $a$ from the first circle, we can proceed to solve for $r$, there will be two values, and we should take the positive one. The other negative value corresponds to the circle that can drawn tangent to the '$1400$' circle but from below.

Can you take it from here ?

Answer 2

Consider an inscribed circle $\gamma$ of centre $C$ and tangency point $P=(x,x^2)$. The tangent at $P$ intersects $x$-axis at $M=(x/2,0)$ (easy to prove property of a parabola). If $H$ and $K$ are the projections of $P$ on cartesian axes, triangles $PHM$ and $PKC$ are similar (because $CP\perp PM$), hence: $$ d:{x\over2}=r:\sqrt{x^4+{x^2\over4}}=x:x^2, $$ where $d=CK$ and $r=CP$ is the radius of $\gamma$. It follows that: $$ d={1\over2} \quad\text{and}\quad r=\sqrt{x^2+{1\over4}}. $$ Hence: $$OC=x^2+{1\over2}=r^2+{1\over4}.$$

If we have a larger circle $\gamma'$, tangent to the parabola and to $\gamma$, with center $O'$ and radius $r'$, we have then:

$$OC'=r'^2+{1\over4}.$$

But $CC'=r+r'$, that is: $$ r'^2-r^2=r'+r, \quad\text{whence:}\quad r'-r=1. $$

If the equation of the parabola were instead $y=ax^2$, we would obviously get: $$r'-r={1\over a}.$$