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Which one is the imaginary circle out of the given cases ? 

Option: 1

g^2 = f^2 = c


Option: 2

g^2 = f^2 = 3 c


Option: 3

g^2 = 2 f^2 = c


Option: 4

g^2=\frac{1}{2}f^2=\frac{1}{4}c


Answers (1)

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As we have learned

Imaginary circle -

g^{2}+f^{2}-c< 0

 

- wherein

The equation of  circle is

x^{2}+y^{2}+2gx+2fy+c= 0 .
 

 

 Here c = 4g^2 , f^2 = 2 g^2

Thus g^{2}+f^{2}-c = g^2 +2 g^2 - 4 g^2 = - g^2

 

 

 

 

Posted by

SANGALDEEP SINGH

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