Question

The set of values of $a$ for which the point$(a-1,a+1)$ lies outside the circle $x^2+y^2=8$ and inside the circle $x^2+y^2-12x+12y-62=0$ is

• A. $(-3\sqrt{2},-\sqrt{3})\cup (\sqrt{3},3\sqrt{2})$
• B. $(-2\sqrt{2},-\sqrt{3})\cup (\sqrt{3},3\sqrt{2})$
• C. $(-3\sqrt{2},\sqrt{3})\cup (\sqrt{3},3\sqrt{2})$
• D. $(-3\sqrt{2},-\sqrt{3})\cup (-\sqrt{3},3\sqrt{2})$

Answer 1

The correct option is A $(-3\sqrt{2},-\sqrt{3})\cup (\sqrt{3},3\sqrt{2})$

Point $(a-1,a+1)$ lies outside the circle $x^2+y^2=8$
$\Rightarrow (a-1{)}^2+(a+1{)}^2-8>0\Rightarrow a^2-3>0a\in (-\infty ,-\sqrt{3})\cup (\sqrt{3}.\infty )\cdots \left(1\right)$

Point $(a-1,a+1)$ lies inside the circle $(x-6{)}^2+(y+6{)}^2-134=0$
$\Rightarrow (a-7{)}^2+(a+7{)}^2-134<0$
$\Rightarrow a^2-18<0$
$\Rightarrow a\in (-3\sqrt{2},3\sqrt{2})\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$

$a\in (-3\sqrt{2},-\sqrt{3})\cup (\sqrt{3},3\sqrt{2})$