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the equation of a circle whose area is 4*3.14 lying in the 1st quadrant and touching both the axis is 1) (x-1)^2 + (y-1)^2=16 2) (x-2)^2 + (y+2)^2=4 3) (x+2)^2 + (y+2)^2=4 4) (x-2)^2 + (y-2)^2=4

the equation of a circle whose area is 4*3.14 lying in the 1st quadrant and touching both the axis is


1) (x-1)^2 + (y-1)^2=16


2) (x-2)^2 + (y+2)^2=4


3) (x+2)^2 + (y+2)^2=4


4) (x-2)^2 + (y-2)^2=4

Grade:12

2 Answers

Sanjeev Malik IIT BHU
101 Points
10 years ago

as circle is in 1st quadrant,thus the coordinates of circle are positive.

for area = 4*3.14, radius =2

so clearly visibl option is 4.

sushant singh
66 Points
10 years ago

hey rani ..

 if the circle is touching both the axis i.e

area = 3.14 x r

thus, 3.14 x r2  = 4 x 3.14

 r= 2 (as it is in first quadrant)

Thus the answer (4) is correct...

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