Consider the region R = {( x , y ) in mathbb{ | vedclass

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(A) Let the circle be $(x-h)^2 + y^2 = r^2$,where the center is $(h, 0)$ and radius is $r$. Since the circle is contained in $R$,it must be tangent to

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$1.50, 2$

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(A) Let the circle be $(x-h)^2 + y^2 = r^2$,where the center is $(h, 0)$ and radius is $r$. Since the circle is contained in $R$,it must be tangent to the parabola $y^2 = 4-x$ at some point $(\alpha, \beta)$.Substituting $y^2 = 4-x$ into the circle equation: $(x-h)^2 + 4-x = r^2$,which simplifies to $x^2 - (2h+1)x + (h^2+4-r^2) = 0$.For tangency,the discriminant must be zero: $D = (2h+1)^2 - 4(h^2+4-r^2) = 0$.$4h^2 + 4h + 1 - 4h^2 - 16 + 4r^2 = 0 \Rightarrow 4h + 4r^2 = 15 \Rightarrow h = \frac{15-4r^2}{4}$.Also,the circle must be contained in $x \geq 0$,so the leftmost point $h-r \geq 0 \Rightarrow h \geq r$.Substituting $h$: $\frac{15-4r^2}{4} \geq r \Rightarrow 15-4r^2 \geq 4r \Rightarrow 4r^2 + 4r - 15 \leq 0$.Solving $4r^2 + 4r - 15 = 0$: $r = \frac{-4 \pm \sqrt{16 - 4(4)(-15)}}{8} = \frac{-4 \pm \sqrt{256}}{8} = \frac{-4 \pm 16}{8}$.Since $r > 0$,$r = \frac{12}{8} = 1.5$.For $r = 1.5$,$h = \frac{15 - 4(2.25)}{4} = \frac{15-9}{4} = 1.5$.The circle is $(x-1.5)^2 + y^2 = 2.25$. Intersection with $y^2 = 4-x$: $(x-1.5)^2 + 4-x = 2.25 \Rightarrow x^2 - 3x + 2.25 + 4 - x = 2.25 \Rightarrow x^2 - 4x + 4 = 0 \Rightarrow (x-2)^2 = 0 \Rightarrow \alpha = 2$.

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