Question

Let $C_1$ and $C_2$ be the centres of the circles $x^2+y^2-2x-2y-2=0$ and $x^2+y^2-6x-6y+14=0$ respectively. If $P$ and $Q$ are the points of intersection of these circles, then the area (in sq. units) of the quadrilateral $P{C}_{1}Q{C}_2$ is :

• A. $4$
• B. $6$
• C. $8$
• D. $9$

Answer 1

The correct option is A $4$

Given circle are
$x^2+y^2-2x-2y-2=0$ and $x^2+y^2-6x-6y+14=0$
$C_1=(1,1),r_1=2C_2=(3,3),r_2=2$


As $P$ lies on the circle, so
$P{C}_1=2,P{C}_2=2C_1{C}_2=\sqrt{4+4}=2\sqrt{2}\Rightarrow (P{C}_1{)}^2+(P{C}_2{)}^2=(C_1{C}_2{)}^2$

$\therefore △C_{1}P{C}_2$ is a right angle triangle

So, the area of quadrilateral $P{C}_{1}Q{C}_2$
$=2(\frac{1}{2}\times P{C}_1\times P{C}_2)=2(\frac{1}{2}\times 2\times 2)=4\text{sq. units}$