Question

If the tangents $PA$ and $PB$ are drawn from the point $P(-1,2)$ to the circle $x^2+y^2+x-2y-3=0$ and $C$ is the center of the circle, then the area of the quadrilateral $PACB$ is

• A. $4$
• B. $16$
• C. Does not exists
• D. $8$

Answer 1

The correct option is C Does not exists

The given circle is $S:x^2+y^2+x-2y-3=0$

Since at point $P(-1,2)$ $S_{(-1,2)}=1+4-1-4-3=-3<0$

the point $P(-1,2)$ lies inside the circle.

Consequently, the tangents from the point $P(-1,2)$ to the circle does not exits.

Thus, the quadrilateral $PACB$ cannot be formed.