Question

$PA$ and $PB$ are tangents drawn from $P(2,3)$ to the circle $x^2+y^2+4x+2y+1=0$. If $C$ is the centre of the circle and $PACB$ is a cyclic quadrilateral then the circumcentre of the $\Delta PAB$ is

• A. $(0,1)$
• B. $(0,-1)$
• C. $(-2,1)$
• D. $(2,1)$

Answer 1

Answer: (A)

$(0,1)$

Given Quadrilateral $PACB$ is cyclic. i.e. $P,A,C$ and $B$ lie on a circle.

Circle passing through points $A$ and $B$ also passes through point $C$, because $PACB$ is cyclic quadrilateral.

By $\Delta$ PAC, $<\angle PAC={90}^0$

Then point $P(2,3)$ and $C(-2,-1)$ are extremities of diameter of circle passing through $P,A,B$. So, circumcentre of the circle is mid-point of $P$ and $C$.

$\therefore$ Center $=(0,\frac{2}{2})=(0,1)$