Question

How do you find the radius of a circle with the equation ${x^2} + {y^2} + 4x - 8y + 13 = 0$ ?

Answer 1

Hint: In this question, we are given an equation of circle and we have to find the radius of the circle. The standard form of the equation of a circle is ${(x - h)^2} + {(y - k)^2} = {r^2}$ where $(h,k)$ are the coordinates of the centre of the circle and $r$ is the radius of the circle (radius is defined as the distance between the centre of the circle and any point on the boundary of the circle). Thus to find the radius of the circle we will convert the given equation to the standard form and then compare both of them.

Complete step by step solution:
We are given that ${x^2} + {y^2} + 4x - 8y + 13 = 0$
We know that standard form of the equation of a circle is ${(x - h)^2} + {(y - k)^2} = {r^2}$
So we will add and subtract some terms in the given question to convert them into standard form –
$
  {x^2} + 4x + 4 - 4 + {y^2} - 8y + 16 - 16 + 13 = 0 \\
   \Rightarrow {x^2} + 4x + {(2)^2} + {y^2} - 8y + {(4)^2} + 13 - 4 - 16 = 0 \;
 $
We know that ${a^2} + {b^2} - 2ab = {(a - b)^2}$ and ${a^2} + {b^2} + 2ab = {(a + b)^2}$ , using these two identities in the obtained equation, we get –
$
   \Rightarrow {(x + 2)^2} + {(y - 4)^2} + 13 - 20 = 0 \\
   \Rightarrow {(x + 2)^2} + {(y - 4)^2} = 7 \;
 $
Now, on comparing the obtained equation with the standard form, we get –
$
  {r^2} = 7 \\
   \Rightarrow r = \pm \sqrt 7 \;
 $
Length can never be negative, so we reject the negative value.
Hence the radius of a circle with the equation ${x^2} + {y^2} + 4x - 8y + 13 = 0$ is $\sqrt 7 $ .
So, the correct answer is “$\sqrt 7 $”.

Note: There are various ways to write the equation of a circle, but the standard way is ${(x - h)^2} + {(y - k)^2} = {r^2}$ . After converting the given equation into the standard way, we can also use it to draw its graph as we will get the coordinates of the centre and the radius of the circle. The coordinates of the centre of this circle is $( - 2,4)$ .