If A_1, A_2, A_3 are the areas of the ellipse | vedclass

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(C) The given equation of the ellipse is $x^2+4y^2-4=0$, which can be written as $\frac{x^2}{4}+\frac{y^2}{1}=1$. Here, $a^2=4$ and $b^2=1$, so $a=2$

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$7$

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(C) The given equation of the ellipse is $x^2+4y^2-4=0$, which can be written as $\frac{x^2}{4}+\frac{y^2}{1}=1$. Here, $a^2=4$ and $b^2=1$, so $a=2$ and $b=1$. $A_1$ is the area of the ellipse, given by $A_1 = \pi ab = \pi(2)(1) = 2\pi$. The equation of the director circle is $x^2+y^2=a^2+b^2 = 4+1=5$. Thus, $A_2$ is the area of the director circle, $A_2 = \pi(r^2) = 5\pi$. The equation of the auxiliary circle is $x^2+y^2=a^2 = 4$. Thus, $A_3$ is the area of the auxiliary circle, $A_3 = \pi(r^2) = 4\pi$. Finally, $A_2+A_3-A_1 = 5\pi+4\pi-2\pi = 7\pi$.

If $A_1, A_2, A_3$ are the areas of the ellipse $x^2+4y^2-4=0$, its director circle, and its auxiliary circle respectively, then $A_2+A_3-A_1=$ (in $\pi$)

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