Question

Let $C:x^2+y^2-x-y-6=0$ and point $P(a-1,a+1)$ lies inside the circle $C$. If the line $x+y-2=0$ divides the circle in two segments, then

• A. $P$ to lie in the larger segment of the intersection if $a\in (-1,1)$
• B. $P$ to lie in the larger segment of the intersection if $a\in (1,2)$
• C. $P$ to lie in the smaller segment of the intersection if $a\in (1,2)$
• D. $P$ to lie in the smaller segment of the intersection if $a\in (-1,1)$

Answer 1

Answer: (A), (C)

$P$ to lie in the smaller segment of the intersection if $a\in (1,2)$

Given qquation of the circle is
$C:x^2+y^2-x-y-6=0$
Centre $c=(\frac{1}{2},\frac{1}{2})$
Radius $r=\sqrt{\frac{13}{2}}$

$P(a-1,a+1)$ lies inside the given circle, so
$(a-1{)}^2+(a+1{)}^2-a+1-a-1-6<0\Rightarrow a^2-a-2<0\Rightarrow (a+1)(a-2)<0\Rightarrow a\in (-1,2)\cdots \left(1\right)$

Now, we know that the centre of the circle always lies in the major segment,
$L\left(c\right)=1-2=-1<0$
So, $P$ lies in the major segment when
$L\left(P\right)<0\Rightarrow a-1+a+1-2<0\Rightarrow a<1$
Using equation $\left(1\right)$
When $a\in (-1,1)$ point $P$ lies in the major segment.
When $a\in (1,2)$ point $P$ lies in the minor segment.
For $a=1$ point $P$ lies on the line