Question

Match elements of list I correctly with elements of list II:

For the circle $S\equiv x^2+y^2+x-y-2=0$, the point

List I	List II
A) $(-2,1)$ lies	1) on the circle
B) $(2,-1)$ lies	2) outside the circle
C) $(0,1)$ lies	3) on the tangent at $(1,0)$ to $S$
D) $(2,3)$ lies	4) inside the circle $S$

The correct matching order for $A,B,C,D$ is

• A. $1234$
• B. $2143$
• C. $3214$
• D. $1243$

Answer 1

The correct option is D $1243$

$S\equiv x^2+y^2+x-y-2=0$

Center of the circle $C_1\equiv (-\frac{1}{2},\frac{1}{2})$

For any point to lie inside/outside or on the circle, it should follow these conditions:

If $S_p>0$, then point $P$ lies outside the circle $S$

If $S_p<0$, then point $P$ lies inside the circle $S$,

If $S_p=0$, then point $P$ lies on the circle $S$.

Tangent to the circle $S$ at point $(x_1,y_1)$ is

$T:x{x}_1+y{y}_1+\frac{x+x_1}{2}-\frac{y+y_1}{2}-2=0$

Tangent at point $(1,0)$ is

$T:x+\frac{x+1}{2}-\frac{y}{2}-2=0$

$\Rightarrow T:3x-y-3=0$

Now,

A) Point $P(-2,1)$

$S_p=(-2{)}^2+(1{)}^2-2-1-2=0$

$T_p=3(-2)-1-3=-10<0$

So, point $P(-2,1)$ lies on the circle and does not lie on tangent at $(1,0)$ to $S$. $\to \left(1\right)$

B) Point $(2,-1)$

$S_p=(2{)}^2+(-1{)}^2+2+1-2=6>0$

$T_p=3\left(2\right)+1-3=4>0$

So, point $P(2,-1)$ lies outside the circleand does not lie on tangent at $(1,0)$ to $S$. $\to \left(2\right)$

C) Point $(0,1)$

$S_p=(0{)}^2+(1{)}^2+0-1-2=-2<0$

$T_p=3\left(0\right)-1-3=-4<0$

So, point $P(0,1)$ lies inside the circleand does not lie on tangent at $(1,0)$ to $S$. $\to \left(4\right)$

D) Point $P(2,3)$

$S_p=2^2+3^2+2-3-2=10>0$

$T_p=3\left(2\right)-3-3=0$

So, point $P(2,3)$ lies outside the circleand also lies on tangent at $(1,0)$ to $S$. $\to \left(3\right)$

$\therefore$ Correct matching order is A(1) B(2) C(4) D(3)

Hence, the best matching option is D.