
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 square units) of the quadrilateral P${C_1}$Q${C_2}$ is:
$
A.\;8 \\
B.\,6 \\
C.\;9 \\
D.\;4 \\
$
Answer
476.4k+ views
Hint: In order to solve this problem we need to know that area of rhombus $\dfrac{1}{2}{d_1} \times {d_2}$. We need to find the radius of the circle and then we need to apply the formula of the area of the rhombus after finding the length of its diagonals. Doing this will solve your problem and will give you the right answer.
Complete step-by-step answer:
The diagram for this problem can be drawn as:
As it is given to us that ${C_1}$ and ${C_2}$ are the centres of the given circles and also that P and Q are the points of intersection of these circles and hence we can say that,
${C_1}$=$\left( {\dfrac{2}{2},\dfrac{2}{2}} \right) = \left( {1,1} \right)$
Similarly,
${C_2}$=$\left( {\dfrac{6}{2},\dfrac{6}{2}} \right) = \left( {3,3} \right)$
Now our Radius ${r_1} = \sqrt {{1^2} + {1^2} + 2} = \sqrt 4 = 2$
Similarly Radius ${r_2} = 2$ and hence
We can say that the quadrilateral is Rhombus since all side are equal.
We see that P${C_1}$, P${C_2}$, Q${C_1}$, Q${C_2}$ are radii of two circles
And hence, P${C_1}$=${r_1}$ =2
And diagonals bisect at O= $\left( {\dfrac{{1 + 3}}{2},\dfrac{{1 + 3}}{2}} \right)$ =(2,2)
Now since diagonals are perpendicular in Rhombus and hence
$ \Rightarrow P{C_1}^2 = P{O^2} + O{C_1}^2$
And hence on putting the value, we have
$
\Rightarrow P{O^2} = {\left( 2 \right)^2} - {\left( {\sqrt 2 } \right)^2} \\
\Rightarrow PO = \sqrt 2 \\
$
$ \Rightarrow PQ = {d_1} = 2\sqrt 2 ,\;\;{C_1}{C_2} = {d_2} = 2\sqrt 2 $
Now Area of Rhombus=$\dfrac{1}{2}{d_1} \times {d_2}$
And hence on putting the value, we have
Area of Rhombus= $\dfrac{1}{2} \times 2\sqrt 2 \times 2\sqrt 2 $
And hence on doing the multiplication, we have
Area of Rhombus=4 square unit
Note: In this type of question first of all we have to find the value of radius and with the help of that we can identify the geometry of the figure and then we have to find the measurement of the diagonal and finally with the help of that we can Area of the required geometry. Knowing this and proceeding like this you will get the right answer.
Complete step-by-step answer:
The diagram for this problem can be drawn as:

As it is given to us that ${C_1}$ and ${C_2}$ are the centres of the given circles and also that P and Q are the points of intersection of these circles and hence we can say that,
${C_1}$=$\left( {\dfrac{2}{2},\dfrac{2}{2}} \right) = \left( {1,1} \right)$
Similarly,
${C_2}$=$\left( {\dfrac{6}{2},\dfrac{6}{2}} \right) = \left( {3,3} \right)$
Now our Radius ${r_1} = \sqrt {{1^2} + {1^2} + 2} = \sqrt 4 = 2$
Similarly Radius ${r_2} = 2$ and hence
We can say that the quadrilateral is Rhombus since all side are equal.
We see that P${C_1}$, P${C_2}$, Q${C_1}$, Q${C_2}$ are radii of two circles
And hence, P${C_1}$=${r_1}$ =2
And diagonals bisect at O= $\left( {\dfrac{{1 + 3}}{2},\dfrac{{1 + 3}}{2}} \right)$ =(2,2)
Now since diagonals are perpendicular in Rhombus and hence
$ \Rightarrow P{C_1}^2 = P{O^2} + O{C_1}^2$
And hence on putting the value, we have
$
\Rightarrow P{O^2} = {\left( 2 \right)^2} - {\left( {\sqrt 2 } \right)^2} \\
\Rightarrow PO = \sqrt 2 \\
$
$ \Rightarrow PQ = {d_1} = 2\sqrt 2 ,\;\;{C_1}{C_2} = {d_2} = 2\sqrt 2 $
Now Area of Rhombus=$\dfrac{1}{2}{d_1} \times {d_2}$
And hence on putting the value, we have
Area of Rhombus= $\dfrac{1}{2} \times 2\sqrt 2 \times 2\sqrt 2 $
And hence on doing the multiplication, we have
Area of Rhombus=4 square unit
Note: In this type of question first of all we have to find the value of radius and with the help of that we can identify the geometry of the figure and then we have to find the measurement of the diagonal and finally with the help of that we can Area of the required geometry. Knowing this and proceeding like this you will get the right answer.
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