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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

rani j , 14 Years ago
Grade 12
anser 2 Answers
Sanjeev Malik IIT BHU

Last Activity: 14 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

Last Activity: 14 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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