Question

If $A=\{(x,y)\mid x^2+y^2\le 4\}$ and $B=\{(x,y)\mid {(x-3)}^2+y^2\le 4\}$ and the point $P(a,a-\frac{1}{2})$ belongs to the set $B-A$, then the set of possible real values of $a$ is:

• A. $(\frac{1+\sqrt{31}}{4},\frac{7+\sqrt{7}}{4}]$
• B. $[\frac{7-\sqrt{7}}{4},\frac{1+\sqrt{31}}{4})$
• C. $(\frac{1-\sqrt{31}}{4},\frac{7-\sqrt{7}}{4}]$
• D. None of the above

Answer 1

Answer: (A)

$(\frac{1+\sqrt{31}}{4},\frac{7+\sqrt{7}}{4}]$

$A$ denotes the points inside the circle of radius $2$ and origin $(0,0)$.
$B$ denotes the points inside the circle of radius $2$ and origin $(3,0)$.
The point $P$ lies inside $B$ and outside $A$.
So the point $P$ should satisfy
$a^2+{(a-\frac{1}{2})}^2-4>0\Rightarrow a\in (-\infty ,\frac{1-\sqrt{31}}{4})\cup (\frac{1+\sqrt{31}}{4},\infty )$
${(a-3)}^2+{(a-\frac{1}{2})}^2-4\le 0\Rightarrow a\in [\frac{7-\sqrt{7}}{4},\frac{7+\sqrt{7}}{4}]$
$\Rightarrow a\in (\frac{1+\sqrt{31}}{4},\frac{7+\sqrt{7}}{4}]$