Question:
<p>The radius of the smallest circle which touches the parabolas <span class="math-tex">\(y=x^{2}\)</span> +2 and <span class="math-tex">\(x=y^{2}+2\)</span> is</p>
<p style="display:inline"><span class="math-tex">\(\frac{7 \sqrt{2}}{2}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{7 \sqrt{2}}{16}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{7 \sqrt{2}}{8}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{7 \sqrt{2}}{4}\)</span></p>
Step-by-Step Solution
Key Concept: The smallest circle touching two curves symmetric about y=x has a diameter equal to the shortest distance between them, occurring where the tangents are parallel to the line of symmetry.
<p><img src="https://media-mycbseguide.s3.amazonaws.com/images/question_images/1775823589-4v7udp.jpg" style="height:259px; width:250px" /><br />
At point <span class="math-tex">$A$</span> and <span class="math-tex">$B$</span> tangent must be parallel to <span class="math-tex">$y=x$</span>, So slope of the tangent <span class="math-tex">$=1$</span><br />
For <span class="math-tex">$y=x^{2}+2 \Rightarrow \frac{d y}{d x}=2 x=1$</span><br />
<span class="math-tex">$\Rightarrow x=\frac{1}{2}, y=\frac{9}{4} \Rightarrow B\left(\frac{1}{2}, \frac{9}{4}\right)$</span><br />
By symmetry, <span class="math-tex">$A\left(\frac{9}{4}, \frac{1}{2}\right)$</span><br />
<span class="math-tex">$A B=\sqrt{\left(\frac{7}{4}\right)^{2}+\left(-\frac{7}{4}\right)^{2}}=\frac{7 \sqrt{2}}{4}$</span><br />
Radius <span class="math-tex">$r=\frac{A B}{2}=\frac{7 \sqrt{2}}{8}$</span></p>
Correct Answer: C