Maths Q11 – JEE-MAIN 2025

The radius of the smallest circle which touches the parabolas

y = x^{2} + 2

and

x = y^{2} + 2

Clue 1 of 3: Symmetry and Center

Observe the symmetry of the two parabolas with respect to the line $y = x$ . The center of the smallest circle touching both parabolas must lie on this line of symmetry.

🥷

Ninja Strategy•Diameter vs. Radius Check

Recognize that the minimum distance between the parabolas is the circle's diameter, and then carefully divide by two to find the radius, eliminating options that represent the diameter or incorrect multiples.

🧠 Full Solution Path

Step 1: Identify Symmetry and Closest Points✦ Active

The two parabolas $y = x^{2} + 2$ and $x = y^{2} + 2$ are symmetric with respect to the line $y = x$ . The smallest circle touching both parabolas will have its center on $y = x$ , and its diameter will be the minimum distance between the two parabolas. The points of tangency will be where the common normal is perpendicular to $y = x$ , meaning its slope is $- 1$ .

For the parabola $y = x^{2} + 2$ , the derivative is $\frac{d y}{d x} = 2 x$ . The slope of the normal at a point $(x, y)$ is $m_{n} = - \frac{1}{2 x}$ . Setting $m_{n} = - 1$ gives:

- \frac{1}{2 x} = - 1 ⟹ 2 x = 1 ⟹ x = \frac{1}{2}

Substitute $x = \frac{1}{2}$ into $y = x^{2} + 2$ to find the point $P_{1}$ :

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

So, $P_{1} = (\frac{1}{2}, \frac{9}{4})$ . By symmetry, the corresponding point $P_{2}$ on the parabola $x = y^{2} + 2$ is $P_{2} = (\frac{9}{4}, \frac{1}{2})$ .

Step 2: Calculate the Minimum Distance (Diameter)○ Expand

The minimum distance between the two parabolas is the distance between $P_{1}$ and $P_{2}$ . This distance will be the diameter of the smallest circle.

D = \sqrt{{(\frac{9}{4} - \frac{1}{2})}^{2} + {(\frac{1}{2} - \frac{9}{4})}^{2}}

Simplify the terms inside the square root:

D = \sqrt{{(\frac{9}{4} - \frac{2}{4})}^{2} + {(\frac{2}{4} - \frac{9}{4})}^{2}} = \sqrt{{(\frac{7}{4})}^{2} + {(- \frac{7}{4})}^{2}}

Calculate the distance:

D = \sqrt{\frac{49}{16} + \frac{49}{16}} = \sqrt{\frac{98}{16}} = \sqrt{\frac{49 \times 2}{16}} = \frac{7 \sqrt{2}}{4}

💡 Teacher's Secret Hint

Remember that the distance between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is $\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$ .

Step 3: Determine the Radius○ Expand

The radius $R$ of the smallest circle is half of the diameter $D$ .

R = \frac{D}{2} = \frac{1}{2} \cdot \frac{7 \sqrt{2}}{4} = \frac{7 \sqrt{2}}{8}

💡 Teacher's Secret Hint

Double-check if the question asks for diameter or radius to avoid common errors.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Maths Question 11 – JEE-MAIN 2025

Clue 1 of 3: Symmetry and Center

🧠 Full Solution Path