Animated Solution for Mathematics - Conic Sections: The radius of the smallest circle which touches the parabolas $y=x^2+2$ and $x=y^2+2$ is

Imagine standing in a coordinate plane, looking at two graceful curves. The first, $y=x^2+2$, is a classic upward-opening parabola, its vertex resting comfortably at $(0,2)$.

The second, $x=y^2+2$, is its sibling, opening to the right with its vertex at $(2,0)$. If you swap the $x$ and $y$ variables in the first equation, you get the second.

They are perfect mirror images, dancing around the line $y=x$. This symmetry is not just a visual treat; it is the key to unlocking the entire problem.

We are tasked with finding the smallest circle that touches both of these curves. Imagine a tiny, perfect marble being squeezed between these two parabolas.

To make the circle as small as possible, we must place it in the narrowest gap between the curves. Geometrically, this minimum distance lies along the common normal—a line that is perpendicular to both curves at the points of contact.

Because our parabolas are symmetric about the line $y=x$, this common normal must be perpendicular to the line $y=x$. Since the slope of $y=x$ is $1$, the slope of our normal must be $-1$. Consequently, the tangents at the points of contact must be parallel to the line $y=x$, meaning their slope must also be $1$.

The Tool of Precision

Now, we turn to calculus to find exactly where these points of contact lie. For the parabola $y=x^2+2$, the slope of the tangent at any point is given by the derivative:

We know the tangent must have a slope of $1$ to be parallel to our line of symmetry. So, we set the derivative equal to $1$:

This is the $x$-coordinate of our first point of contact, let's call it $A$. To find the $y$-coordinate, we plug $x=\frac{1}{2}$ back into the original equation:

So, point $A$ is $\left(\frac{1}{2},\frac{9}{4}\right)$.

We could repeat this process for the second parabola, but why work harder when we can work smarter? By the principle of symmetry across $y=x$, the closest point on the second parabola, point $B$, is simply the reflection of point $A$.

We just swap the coordinates:

Now, we have both points of contact. The diameter of our smallest circle is simply the distance between $A$ and $B$. Using the distance formula:

Simplifying the terms inside:

This is the diameter of our circle. The question asks for the radius, which is half the diameter:

We have arrived at our destination: $\frac{7\sqrt{2}}{8}$. It is a beautiful result, born from the elegant interplay of symmetry and calculus.

Remember, in JEE Advanced, the most complex-looking problems often yield to the simplest geometric insights. Never lose sight of the symmetry; it is often the shortest path to the truth.