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

Question

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

Sanjeev Malik IIT BHU · Answer

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 · Answer

hey rani .. if the circle is touching both the axis i.e area = 3.14 x r 2 thus, 3.14 x r 2 = 4 x 3.14 r= 2 (as it is in first quadrant) Thus the answer (4) is correct...

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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

2 Answers

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