To find the radius of the circle given by the equation [tex]\( x^2 + y^2 + 8x - 6y + 21 = 0 \)[/tex], we need to convert it into the standard form of a circle's equation, which is [tex]\( (x-h)^2 + (y-k)^2 = r^2 \)[/tex].
1. Reorganize the terms:
Group the [tex]\( x \)[/tex] and [tex]\( y \)[/tex] terms together and move the constant term to the other side of the equation:
[tex]\[
x^2 + 8x + y^2 - 6y = -21
\][/tex]
2. Complete the square for the [tex]\( x \)[/tex] terms:
- Take the coefficient of [tex]\( x \)[/tex], which is 8, divide it by 2 to get 4, and then square it to get 16.
- Add and subtract this square inside the equation:
[tex]\[
x^2 + 8x + 16 + y^2 - 6y = -21 + 16
\][/tex]
3. Complete the square for the [tex]\( y \)[/tex] terms:
- Take the coefficient of [tex]\( y \)[/tex], which is -6, divide it by 2 to get -3, and then square it to get 9.
- Add and subtract this square inside the equation:
[tex]\[
x^2 + 8x + 16 + y^2 - 6y + 9 = -21 + 16 + 9
\][/tex]
4. Write both groups as perfect squares:
- The [tex]\( x \)[/tex] terms complete to [tex]\( (x + 4)^2 \)[/tex]
- The [tex]\( y \)[/tex] terms complete to [tex]\( (y - 3)^2 \)[/tex]
- The right side simplifies to 4:
[tex]\[
(x + 4)^2 + (y - 3)^2 = 4
\][/tex]
In the final equation, [tex]\( (x + 4)^2 + (y - 3)^2 = 4 \)[/tex], the standard form of a circle [tex]\((x-h)^2 + (y-k)^2 = r^2\)[/tex], the radius [tex]\( r \)[/tex] is the square root of the constant on the right-hand side.
So, the radius [tex]\( r \)[/tex] is [tex]\( \sqrt{4} = 2 \)[/tex] units.
Thus, the radius of the circle is 2 units.
