The radius of the smallest circle which touches the parabolas and is [2025]

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Solution

Q.
The radius of the smallest circle which touches the parabolas $y = x^{2} + 2$ and $x = y^{2} + 2$ is [2025]

1 $\frac{7 \sqrt{2}}{2}$

2 $\frac{7 \sqrt{2}}{16}$

3 $\frac{7 \sqrt{2}}{4}$

4 $\frac{7 \sqrt{2}}{8}$

Ans.
(4)

The given parabolas are symmetric about the line y = x.

Tangents at A and B must be parallel to line y = x, so slope of the tangents = 1, which is minimum.

${(\frac{d y}{d x})}_{\min A} = 1 = {(\frac{d y}{d x})}_{\min B}$

For point B, $y = x^{2} + 2$

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

When $x = \frac{1}{2} \Rightarrow y = \frac{9}{4}$

$∴ Point B = (\frac{1}{2}, \frac{9}{4})$

Similarly, point $A = (\frac{9}{4}, \frac{1}{2})$

$A B = \sqrt{{(\frac{1}{2} - \frac{9}{4})}^{2} + {(\frac{9}{4} - \frac{1}{2})}^{2}} = \sqrt{\frac{98}{16}} = \frac{7 \sqrt{2}}{4}$

Radius = $\frac{\frac{7 \sqrt{2}}{4}}{2} = \frac{7 \sqrt{2}}{8}$ .

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