Question

The radius of the smallest circle which touches the parabolas $ y = x^2 + 2 $ and $ x = y^2 + 2 $ is

• A. \( \frac{7\sqrt{2}}{2} \)
• B. \( \frac{7\sqrt{2}}{16} \)
• C. \( \frac{7\sqrt{2}}{4} \)
• D. \( \frac{7\sqrt{2}}{8} \)

Hint:

Use the symmetry of the parabolas about the line \( y = x \) to find the points of tangency and then calculate the distance between them.

Answer 1

The Correct Option is D

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

1. \( \left( \frac{dy}{dx} \right)_{\min A} = 1 = \left( \frac{dy}{dx} \right)_{\min B} \) 

For \( y = x^2 + 2 \), \( \frac{dy}{dx} = 2x \) \( 2x = 1 \) \( x = \frac{1}{2} \) \( y = \left( \frac{1}{2} \right)^2 + 2 = \frac{1}{4} + 2 = \frac{9}{4} \) 

So, point A is \( \left( \frac{1}{2}, \frac{9}{4} \right) \). For \( x = y^2 + 2 \), \( 1 = 2y \frac{dy}{dx} \) \( \frac{dy}{dx} = \frac{1}{2y} \) \( \frac{1}{2y} = 1 \) \( y = \frac{1}{2} \) \( x = \left( \frac{1}{2} \right)^2 + 2 = \frac{1}{4} + 2 = \frac{9}{4} \) 

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

Distance between A and B: \( AB = \sqrt{ \left( \frac{9}{4} - \frac{1}{2} \right)^2 + \left( \frac{1}{2} - \frac{9}{4} \right)^2 } \) \( AB = \sqrt{ 2 \left( \frac{7}{4} \right)^2 } \) \( AB = \frac{7\sqrt{2}}{4} \) 

The radius of the smallest circle is half of the distance AB. Radius = \( \frac{AB}{2} = \frac{7\sqrt{2}}{8} \)