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. $8$
• B. $6$
• C. $9$
• D. $4$

Answer 1

The correct option is D $4$

$C_1=(\frac{2}{2},\frac{2}{2})=(1,1)$

$C_2=(\frac{6}{2},\frac{6}{2})=(3,3)$

Radius $r_1=\sqrt{1^2+1^2+2}=2$

Similarly radius $r_2=2$

Here the quadrilateral is rhombus as all sides are equal

As we can see that $P{C}_1,P{C}_2,Q{C}_1,Q{C}_2$ are radii of two circles

Here $P{C}_1=r_1=2$

And diagonals bisect at $O=(\frac{1+3}{2},\frac{1+3}{2})=(2,2)$

$\Rightarrow P{C}_1^2=P{O}^2+O{C}_1^2$ (as diagonals are perpendicular in rhombus)

$\Rightarrow P{O}^2=2^2-{\sqrt{2}}^2$

$\Rightarrow PO=\sqrt{2}$

$\Rightarrow PQ=d_1=2\sqrt{2},C_1{C}_2=d_2=2\sqrt{2}$

$\Rightarrow$Area of rhombus =$\frac{1}{2}{d}_1{d}_2$

$\Rightarrow$Area of rhombus =$\frac{1}{2}\times 2\sqrt{2}\times 2\sqrt{2}$

$\Rightarrow \text{Area of rhombus}=4$