Gently does it

The center of the first circle will lie on the $y$ -axis. Let the center of this circle be $(0,k)$ and its radius be $r$ . Then the equation of this circle will be $x^{2} + (y - k)^{2} = r^{2}$ . To find the two points of tangency between this circle and the parabola $y = x^{2}$ we substitute $y = x^{2}$ into the equation of the circle to find that

$y + (y - k)^{2} = r^{2} \Longrightarrow y^{2} - (2k - 1)y + (k^{2} - r^{2}) = 0$ .

Employing the quadratic formula, we find that

$y = (\frac{1}{2})((2k - 1) \pm \sqrt{(2k - 1)^{2} - 4(k^{2} - r^{2})}) \Longrightarrow y = (\frac{1}{2})((2k - 1) \pm \sqrt{-4k^{2} + 1 + 4r^{2}})$ .

Now by symmetry the $y$ -coordinates of the two points of tangency must be the same, which implies that the discriminant must equal zero. Thus

$-4k + 1 + 4r^{2} = 0 \Longrightarrow k = r^{2} + \frac{1}{4}$ .

The $y$ -coordinates of the points of tangency are then $k - \frac{1}{2} = r^{2} - \frac{1}{4}$ . Now look at the case where the two circles in question are drawn so that they are tangent to one another. Form a right triangle with one side joining their centers, one side down the $y$ -axis from the center of the first circle, and a third side oriented horizontally joining the center of the second circle and the $y$ -axis.

The hypotenuse will then have length $2r$ and the vertical side will have length $r^{2} + \frac{1}{4} - r$ . But since the point of tangency is a vertical distance of $\frac{1}{2}$ below the center of the first circle, we also know that the vertical side of the triangle has length $2 * \frac{1}{2} = 1$ . Thus

$r^{2} - r + \frac{1}{4} = 1 \Longrightarrow 4r^{2} - 4r - 3 = 0 \Longrightarrow (2r - 3)(2r + 1) = 0 \Longrightarrow r = \frac{3}{2}, -\frac{1}{2}$ .

Since $r \gt 0$ we must have $r = \frac{3}{2}$ , and so $a = 3, b = 2$ and $a + b = \boxed{5}$ .

Brian, this is one of the best problems I've come across on Brilliant, in that choosing the RIGHT way to solve this can make a huge difference how difficult it will be to solve this. I'll post my approach to this problem shortly.

Okay, here it is:

First, let the circles have radius $r$ , the first one centered at $(0,ra)$ , and the second one centered at $(rb,0)$ . Because they are tangent, we know that

${ \left( ra-r \right) }^{ 2 }+{ rb }^{ 2 }={ (2r) }^{ 2 }$

From this, we have

$b=\sqrt { 3+2a-{ a }^{ 2 } }$

The tangent point is at

$\left( \dfrac { rb }{ 2 } ,\dfrac { ra+r }{ 2 } \right)$

And the slope of the tangent line is

$\dfrac { rb }{ ra-r }$

Let the equation of the parabola be (the reason for the coefficients will become clear)

$y=kr{ \left( \dfrac { x }{ r } \right) }^{ 2 }=\left( \dfrac { k }{ r } \right) { x }^{ 2 }$

The slope of this parabola is

$2{ \left( \dfrac { k }{ r } \right) }x$

Then we solve this system of equations

$2\left( \dfrac { k }{ r } \right) \left( \dfrac { rb }{ 2 } \right) =\dfrac { rb }{ ra-r }$

$\left( \dfrac { k }{ r } \right) { \left( \dfrac { rb }{ 2 } \right) }^{ 2 }=\dfrac { ra+r }{ 2 }$

Ending up with $a=\dfrac { 5 }{ 3 }$ and $k=\dfrac { 3 }{ 2 }$ (because $r$ drops out)

Thus, for the parabola to have the form $y={ x }^{ 2 }$ ,
$r$ must be equal to $k$ , and we have our answer.

This approach makes careful use of scaling to simplify things.

Let the point of triple intersection be $(t,t^2)$ . The first circle is $x^2+(y-t^2-\frac12)^2=r^2$ . The second circle is $(x-t(2t^2-2r+1))^2+\left(y-r\right)^2=r^2$ . Since both pass through $(t,t^2)$ , we get $4r^2=4t^2+1$ and $(t-t(2t^2-2r+1))^2+(t^2-r)^2=r^2$ , we solve this system of equations to get $r=\frac32$ .

Ujjwal

(I suggest you sketch the figure and write coordinates of the points)

Let, T(h, h^2) = Common point of tangency for the two circles and parabola

A = y-intercept of tangent. Using property of a parabola : MO = OA

Hence, A = (0, -h^2)

M(0, h^2) = Foot of perpendicular from T on axis

C = Intersection of normal at T and axis

Right triangles AMT & MTC are similar

MT = h, MA = 2 h^2 giving MC = 1/2

Triangles TMC & C2NT are congruent

=> C2 = (2h, h^2-1/2)

=> radius = y coordinate of C2 = h^2 - 1/2 - - - - (I)

But radius is also CT

Pythagoras : CT = sqrt(h^2 + 1/4) - - - - (II)

Solve I & II to get h^2 = 2 and R = 3/2

For the circle tangent in the graph $y={x}^{2}$ , we want to find how high it is from its points of tangent. Differentiating $y={x}^{2}$ : $\frac{dy}{dx}=2{x}$ Then the gradient of the normal of the graph would be $\frac{1}{2x}$ Since the normal would pass through the center of the circle, using trigonometry, one could derive that the circle would always be $0.5$ above its points of tangent:

p

Then the relationship between the radius of the circle $(r)$ and the height $(y)$ is $r=\sqrt { 0.25+y }$

For the second circle, we can tell the height would be $r+(something)$ :

p

where that something is $rsin\theta$

It can then be derived from the gradient of the normal of the graph the length of that "something". The height with relation to the radius of the second circle can then be expressed: $y=r+\frac{r}{\sqrt{4y+1}}$ $r=\frac { y }{ 1+\frac { 1 }{ \sqrt { 4y+1 } } }$

Since the radius and the height must be the same $r=\frac { y }{ 1+\frac { 1 }{ \sqrt { 4y+1 } } } =\sqrt { 0.25+y }$ Solving $\frac { y }{ 1+\frac { 1 }{ \sqrt { 4y+1 } } } =\sqrt { 0.25+y }$ gives $y=2$ .

Plugging $y=2$ in to the above formula yields $r=\frac{3}{2}=\frac{a}{b}$ . Therefore, $a+b=\boxed{5}$