The radius of the circle with minimum area th | vedclass

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(D) Let the circle be $x^2 + (y - k)^2 = r^2$. Since it touches $y = |x|$,the distance from $(0, k)$ to $x - y = 0$ is $r$,so $\frac{|-k|}{\sqrt{2}} =

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$\frac{-2 + \sqrt{34}}{2\sqrt{2}}$

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(D) Let the circle be $x^2 + (y - k)^2 = r^2$. Since it touches $y = |x|$,the distance from $(0, k)$ to $x - y = 0$ is $r$,so $\frac{|-k|}{\sqrt{2}} = r$,which gives $k = r\sqrt{2}$.Substituting $x^2 = 4 - y$ into the circle equation: $(4 - y) + (y - k)^2 = r^2$.$4 - y + y^2 - 2ky + k^2 = \frac{k^2}{2}$.$y^2 - (2k + 1)y + (4 + \frac{k^2}{2}) = 0$.For tangency,the discriminant $D = 0$:$(2k + 1)^2 - 4(4 + \frac{k^2}{2}) = 0$.$4k^2 + 4k + 1 - 16 - 2k^2 = 0$.$2k^2 + 4k - 15 = 0$.Solving for $k$ (taking $k > 0$): $k = \frac{-4 + \sqrt{16 - 4(2)(-15)}}{4} = \frac{-4 + \sqrt{136}}{4} = \frac{-4 + 2\sqrt{34}}{4} = \frac{-2 + \sqrt{34}}{2}$.Since $r = \frac{k}{\sqrt{2}}$,$r = \frac{-2 + \sqrt{34}}{2\sqrt{2}}$.

The radius of the circle with minimum area that touches the curve $y = 4 - x^2$ and the lines $y = |x|$ is:

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