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  • What is the nature of the circle with the equation <img alt="x^{2}+y^{2}+4x-6y+13= 0" src="https://entrancecorner.oncodecogs.com/gif.latex?x%5E%7B2%7D&plus;y%5E%7B2%7D&plus;4x-6y&p

What is the nature of the circle with the equation x^{2}+y^{2}+4x-6y+13= 0?

Option: 1

Real circle


Option: 2

Point circle


Option: 3

Imaginary circle


Option: 4

None of the above


Answers (1)

best_answer

The given equation of the circle is x^{2}+y^{2}+4x-6y+13= 0

To determine the nature of the circle, we need to convert the given equation into the standard form of the circle's equation, i.e., 
\left ( x-h \right )^{2}+\left ( y-k \right )^{2}= r^{2}

where (h, k) is the centre of the circle and r is its radius.

So, let's complete the square for x and y terms in the given equation:
x^{2}+y^{2}+4x-6y+13= 0

\Rightarrow \left ( x^{2} +4x+4\right )+\left ( y^{2} -6y+9\right )= -13+4+9 (adding and subtracting constants to complete the square)
\Rightarrow \left ( x+2 \right )^{2}+\left ( y-3 \right )^{2}= 0 (simplifying)

Now, we can see that the given circle's equation is in the standard form with centre (-2, 3) and radius 0.

The centre of the circle is at (-2, 3), which means it is not a point circle.

However, the radius of the circle is 0, which means the circle doesn't exist.

Therefore, the nature of the circle is imaginary (since the radius is imaginary).



 

Posted by

Divya Prakash Singh

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