Q93 / 160 · AP EAMCET 2024 (21 May Shift 1) · Mathematics
If $\mathrm{A}_1, \mathrm{~A}_2, \mathrm{~A}_3$ are the areas of ellipse $x^2+4 y^2-4=0$ its director circle and auxiliary circle respectively, then $\mathrm{A}_2+\mathrm{A}_3-\mathrm{A}_1=$
- A$11 \pi$
- B$3 \pi$
- Correct answerC$7 \pi$
- D$9 \pi$
Solution
$\begin{aligned}
& \text { } x^2+4 y^2-4=0 \\
& \Rightarrow \frac{x^2}{4}+\frac{y^2}{1}=1, \Rightarrow a^2=4, b^2=1
\end{aligned}$
Equation of director circle is $x^2+y^2=a^2+b^2 \Rightarrow x^2+y^2=5$
Equation of Auxiliary circle is $x^2+y^2=a^2 \Rightarrow x^2+y^2=4$ $\mathrm{A}_1=$ Area of ellipse $=\pi a b=2 \pi$ $A_2=$ Area of Director circle $=5 \pi$ $\mathrm{A}_3=$ Area of Auxiliary circle $=4 \pi$ So, $A_2+A_3-A_1=5 \pi+4 \pi-2 \pi=7 \pi$
Equation of director circle is $x^2+y^2=a^2+b^2 \Rightarrow x^2+y^2=5$
Equation of Auxiliary circle is $x^2+y^2=a^2 \Rightarrow x^2+y^2=4$ $\mathrm{A}_1=$ Area of ellipse $=\pi a b=2 \pi$ $A_2=$ Area of Director circle $=5 \pi$ $\mathrm{A}_3=$ Area of Auxiliary circle $=4 \pi$ So, $A_2+A_3-A_1=5 \pi+4 \pi-2 \pi=7 \pi$
Hence, the correct answer is option: 'C'.