The problem is to find the standard form of the equation of a circle given its general form: $x^2 + y^2 - 8x + 4y + 13 = 0$.

1. Problem Description

The problem is to find the standard form of the equation of a circle given its general form: $x^2 + y^2 - 8x + 4y + 13 = 0$.

2. Solution Steps

We need to complete the square for both the $x$ and $y$ terms.

The general equation of a circle is $(x-h)^2 + (y-k)^2 = r^2$, where $(h, k)$ is the center of the circle and $r$ is the radius.

Starting with the given equation: $x^2 + y^2 - 8x + 4y + 13 = 0$.

Group the $x$ and $y$ terms:

$(x^2 - 8x) + (y^2 + 4y) = -13$

To complete the square for the $x$ terms, we take half of the coefficient of $x$, which is $-8/2 = -4$, and square it: $(-4)^2 = 16$.

To complete the square for the $y$ terms, we take half of the coefficient of $y$, which is $4/2 = 2$, and square it: $(2)^2 = 4$.

Add these values to both sides of the equation:

$(x^2 - 8x + 16) + (y^2 + 4y + 4) = -13 + 16 + 4$

Now, rewrite the expressions in parentheses as squared terms:

$(x - 4)^2 + (y + 2)^2 = 7$

This is the standard form of the equation of the circle.

3. Final Answer

The standard form of the equation is $(x - 4)^2 + (y + 2)^2 = 7$.