What is the radius of the circle ${x^2} + {y^2} + 4x + 6y + 13 = 0$?
A. $\sqrt {26} $
B. $\sqrt {13} $
C. $\sqrt {23} $
D. 0

Hint: We will write the given equation in the form ${\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}$. Then compare the equation with ${\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}$ and calculate the value of $r$. The value of $r$ is the radius of the circle.

Formula Used:
The standard formula of a circle is ${\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}$, where $\left( {a,b} \right)$ is the centre and $r$ is radius.

Complete step by step solution:
Given equation is ${x^2} + {y^2} + 4x + 6y + 13 = 0$.
Arrange the terms of the equation:
${x^2} + 4x + {y^2} + 6y + 13 = 0$
Add and subtract 4:
$ \Rightarrow {x^2} + 4x + 4 + {y^2} + 6y + 13 - 4 = 0$
Add and subtract 9:
$ \Rightarrow {x^2} + 4x + 4 + {y^2} + 6y + 9 + 13 - 4 - 9 = 0$
Apply the algebraical identity:
$ \Rightarrow \left( {{x^2} + 4x + 4} \right) + \left( {{y^2} + 6y + 9} \right) + 13 - 4 - 9 = 0$
$ \Rightarrow {\left( {x + 2} \right)^2} + {\left( {y + 3} \right)^2} + 13 - 4 - 9 = 0$
Add and subtract like terms:
$ \Rightarrow {\left( {x + 2} \right)^2} + {\left( {y + 3} \right)^2} + 0 = 0$
$ \Rightarrow {\left( {x + 2} \right)^2} + {\left( {y + 3} \right)^2} = 0$
Now compare the above equation with ${\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}$
${r^2} = 0$
$ \Rightarrow r = 0$
So, the radius of the circle is 0.

Option ‘D’ is correct

Note: We can solve the given by using another method. First we will compare the given equation with ${x^2} + {y^2} + 2gx + 2fy + c = 0$. Then apply the formula $\sqrt {{g^2} + {f^2} - c} $ to calculate the radius of the circle.
Here $g = 2$, $f = 3$, and $c = 13$.
$\sqrt {{g^2} + {f^2} - c}=\sqrt{2^2+3^2-13} = 0 $
The radius is zero.