InterviewSolution
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InterviewSolution
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The radius of a circle, having minimum area, which touches the curve `y=4−x^2` and the lines, `y=|x|` is: (a) `4(sqrt2+1)` (b) `2(sqrt2+1)` (c) `2(sqrt2-1)` (d) `4(sqrt2-1)` |
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Answer» We can create a rough diagram for the curve `y = 4-x^2` and line `y = |x|`. Please refer to video for the figure. From the figure, we can see that x-coordinate of the circle will be `0`. `:.` Curve `y= 4-x^2` will touch the `y`-axis at `y = 4`. Let radius of the circle is `r`. Then, center of the circle will be `(0,4-r)` Also, the distace from the center of the circle to the line `y-x = 0` is `r`. `:. (4-r-0)/(sqrt(1^2+(-1)^2)) = r` `=>4-r = sqrt2r` `=>r = 4/(sqrt2+1)**(sqrt2-1)/(sqrt2-1) = 4(sqrt2-1)` So, option `(d)` is the correct option. |
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