Question

$IfA=\{\left(x,y\right)|x^2+y^2\le 4\}and$
$B=\{\left(x,y\right)|{(x-3)}^2+y^2\le 4\}andthe$
$po\mathrm{int}P\left(a,\frac{1}{2}\right)belongstothesetB-A$ then the set of possible real values of $a$ is

• A. $\left(\frac{1+\sqrt{3}}{4},\frac{7+\sqrt{7}}{4}\right)$
• B. $\left(\frac{7-\sqrt{7}}{4},\frac{1+\sqrt{7}}{4}\right)$
• C. $\left(\frac{1-\sqrt{31}}{4},\frac{7-\sqrt{7}}{4}\right)$
• D. none of these

Answer 1

The correct option is A $\left(\frac{1+\sqrt{3}}{4},\frac{7+\sqrt{7}}{4}\right)$

$x^2+y^2>4$
$a^2+{\left(a-\frac{1}{2}\right)}^2>4$
$2a^2-a-\frac{5}{4}>0$
$a=\frac{1\pm \sqrt{31}}{4}$
$a\in \left(-\infty ,\frac{1-\sqrt{31}}{4}\right)V\left(\frac{1+\sqrt{31}}{4},\infty \right)$
$a\in \left(\frac{1+\sqrt{3}}{4},\frac{7+\sqrt{7}}{4}\right)$