Question

If $(x+\alpha {)}^2+(y+\beta {)}^2=a^2$;
Then: $x^2+y^2$ has

• A. $Extrema:a^2+{\alpha }^2+{\beta }^2+2a\sqrt{{\alpha }^2+{\beta }^2}$
• B. $Extrema:a^2+{\alpha }^2+{\beta }^2-2a\sqrt{(a+\alpha {)}^2+(a+\beta {)}^2}$
• C. $Extrema:a^2+{\alpha }^2+{\beta }^2+2a\sqrt{(a-\alpha {)}^2+(a-\beta {)}^2}$
• D. $Extrema:a^2+{\alpha }^2+{\beta }^2-2a\sqrt{{\alpha }^2+{\beta }^2}$

Answer 1

Answer: (A), (D)

$Extrema:a^2+{\alpha }^2+{\beta }^2-2a\sqrt{{\alpha }^2+{\beta }^2}$
$Extrema:a^2+{\alpha }^2+{\beta }^2+2a\sqrt{{\alpha }^2+{\beta }^2}$

Let $x+\alpha =a\cos \theta$
$y+\beta =a\sin \theta$

$x^2+y^2=(a\cos \theta -\alpha {)}^2+(a\sin \theta -\beta {)}^2$
$=a^2+{\alpha }^2+{\beta }^2-2a\left(\right(\cos \theta \left)\right(\alpha )+(\sin \theta \left)\right(\beta \left)\right)$

$x^2+y^2=a^2+{\alpha }^2+{\beta }^2-2a(\cos \theta \frac{\alpha }{\sqrt{{\alpha }^2+{\beta }^2}}+\sin \theta \frac{\beta }{\sqrt{{\alpha }^2+{\beta }^2}})\times \sqrt{{\alpha }^2+{\beta }^2}$

If we let, $\frac{\alpha }{\sqrt{{\alpha }^2+{\beta }^2}}=\cos \varphi$ and $\frac{\beta }{\sqrt{{\alpha }^2+{\beta }^2}}=\sin \varphi$;

We get
$x^2+y^2=a^2+{\alpha }^2+{\beta }^2-2a\sqrt{{\alpha }^2+{\beta }^2}(\cos \theta \cos \varphi +\sin \theta \sin \varphi )$
$x^2+y^2=a^2+{\alpha }^2+{\beta }^2-2a\sqrt{{\alpha }^2+{\beta }^2}\left(\cos \right(\theta -\varphi \left)\right)$.... since $-1\le \cos (...)\le 1$

$\therefore Max.(x^2+y^2)=a^2+{\alpha }^2+{\beta }^2+2a\sqrt{{\alpha }^2+{\beta }^2}$

$Min.(x^2+y^2)=a^2+{\alpha }^2+{\beta }^2-2a\sqrt{{\alpha }^2+{\beta }^2}$
Options (A) and (D).