JEE Mathematics 2025 Question with Solution

Given: The parabolas are $y = x^2 + 2$ and $x = y^2 + 2$.

Find: The radius of the smallest circle touching both parabolas.

The given parabolas are symmetric about the line $y = x$. Tangents at $A$ and $B$ must be parallel to the line $y = x$, so the slope of the tangents is $1$.

For $y = x^2 + 2$,

$$\frac{dy}{dx} = 2x$$

Since the tangent slope is $1$,

$$2x = 1$$ $$x = \frac{1}{2}$$

Then

$$y = \left(\frac{1}{2}\right)^2 + 2 = \frac{1}{4} + 2 = \frac{9}{4}$$

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

For $x = y^2 + 2$,

$$1 = 2y\frac{dy}{dx}$$ $$\frac{dy}{dx} = \frac{1}{2y}$$

Again, tangent slope is $1$, so

$$\frac{1}{2y} = 1$$ $$y = \frac{1}{2}$$

Then

$$x = \left(\frac{1}{2}\right)^2 + 2 = \frac{1}{4} + 2 = \frac{9}{4}$$

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

Now the distance between $A$ and $B$ is

$$AB = \sqrt{\left(\frac{9}{4} - \frac{1}{2}\right)^2 + \left(\frac{1}{2} - \frac{9}{4}\right)^2}$$ $$AB = \sqrt{2\left(\frac{7}{4}\right)^2}$$ $$AB = \frac{7\sqrt{2}}{4}$$

The radius of the smallest circle is half of $AB$.

$$r = \frac{AB}{2} = \frac{7\sqrt{2}}{8}$$

Therefore, the correct option is D.

Given: The two parabolas are $y = x^2 + 2$ and $x = y^2 + 2$.

Find: The minimum possible radius of a circle touching both parabolas.

Use the symmetry of the two parabolas about the line $y = x$. For the smallest such circle, the points of contact are symmetric and the common tangents at those points are parallel to $y = x$.

So we first locate the points where slope is $1$ on each parabola.

For $y = x^2 + 2$,

$$\frac{dy}{dx} = 2x$$

Setting slope equal to $1$,

$$2x = 1 \Rightarrow x = \frac{1}{2}$$

Hence,

$$y = x^2 + 2 = \frac{1}{4} + 2 = \frac{9}{4}$$

Thus,

$$A = \left(\frac{1}{2}, \frac{9}{4}\right)$$

For $x = y^2 + 2$, differentiate implicitly:

$$1 = 2y\frac{dy}{dx}$$

So,

$$\frac{dy}{dx} = \frac{1}{2y}$$

Again slope is $1$, therefore

$$\frac{1}{2y} = 1 \Rightarrow y = \frac{1}{2}$$

Then,

$$x = y^2 + 2 = \frac{1}{4} + 2 = \frac{9}{4}$$

Thus,

$$B = \left(\frac{9}{4}, \frac{1}{2}\right)$$

Now compute the distance:

$$AB = \sqrt{\left(\frac{9}{4} - \frac{1}{2}\right)^2 + \left(\frac{1}{2} - \frac{9}{4}\right)^2}$$ $$= \sqrt{\left(\frac{7}{4}\right)^2 + \left(-\frac{7}{4}\right)^2}$$ $$= \sqrt{2\cdot \frac{49}{16}} = \frac{7\sqrt{2}}{4}$$

The smallest circle touching both parabolas has diameter $AB$, so its radius is

$$r = \frac{1}{2} \cdot \frac{7\sqrt{2}}{4} = \frac{7\sqrt{2}}{8}$$

Therefore, the radius is $\frac{7\sqrt{2}}{8}$ and the correct option is D.