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The equation of a circle is x^2+y^2-6 x-4 y+12=0 .  Determine the nature of the circle.

 

Option: 1

Real Circle

 


Option: 2

Imaginary Circle

 


Option: 3

Point Circle

 


Option: 4

None of the above


Answers (1)

best_answer

Comparing the given equation with the standard equation of a circle, we get:

\begin{aligned} & 2 g=-6,2 f=-4, c=12 \\ & g^2+f^2-c=(-3)^2+(-2)^2-12=9+4-12=1 \end{aligned}

Since g^2+f^2-c>0  the radius of the circle is real. Therefore, the given circle is a real circle.


 

Posted by

Ritika Jonwal

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