Let C1 and C2 be centres of two circles

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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 circlesAnd hence, P${C_1}$=${r_1}$ =2And 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 haveArea of Rhombus= $\dfrac{1}{2} \times 2\sqrt 2 \times 2\sqrt 2 $ And hence on doing the multiplication, we haveArea 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.

Answer (Detailed Solution Below)

Option 2 : 4

Free

90 Qs. 360 Marks 180 Mins

From question, the equations of the circles are:

x2 + y2 – 2x – 2y – 2 = 0 ----(1)

x2 + y2 – 6x – 6y + 14 = 0 ----(2)

The two circles are intersected orthogonally if 2g1g2 + 2f1 f2 = c1 + c2.

The above equations are intersecting each other orthogonally, because

⇒ 2(1)(3) + 2(1)(3) = 14 – 2

So, the area of quadrilateral pC1 QC2 is given by the formula:

⇒ A = 2 × ar(ΔPC1C2)

∴ A = 4 sq units

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Let C1 and C2 be the centres of the circles x2+y2 2 x 2 y 2=0 and x2+y2 6 x 6 y+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 QC 2 is :A. 4B. 6C. 8D. 9

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Let C1 and C2 be centres of two circles

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