Circle Equation Analysis

The question asks to determine the radius of a given circle from its equation. The equation provided is \(x^2 + y^2 - 4x + 2y - 31 = 0\). To find the radius, we need to compare this equation with the general form of a circle's equation.

The general equation of a circle is given by:

\(x^2 + y^2 + 2gx + 2fy + c = 0\)

Here, the center of the circle is \((-g, -f)\) and the radius \(r\) is calculated using the formula:

\(r = \sqrt{g^2 + f^2 - c}\)

Identifying Coefficients from the Equation

Let's compare the given equation \(x^2 + y^2 - 4x + 2y - 31 = 0\) with the general form \(x^2 + y^2 + 2gx + 2fy + c = 0\). By comparing the coefficients of \(x\), \(y\), and the constant term, we can find the values of \(g\), \(f\), and \(c\).

• Coefficient of \(x\): \(2g = -4\)
• Coefficient of \(y\): \(2f = 2\)
• Constant term: \(c = -31\)

From these comparisons, we can calculate the values of \(g\) and \(f\):

• \(g = \frac{-4}{2} = -2\)
• \(f = \frac{2}{2} = 1\)

Radius Formula Application

Now that we have the values of \(g\), \(f\), and \(c\), we can substitute them into the radius formula:

\(r = \sqrt{g^2 + f^2 - c}\)

Substitute the values:

\(r = \sqrt{(-2)^2 + (1)^2 - (-31)}\)

\(r = \sqrt{4 + 1 + 31}\)

\(r = \sqrt{36}\)

Final Radius Determination

The square root of 36 is 6.

\(r = 6\)

Therefore, the radius of the circle is 6 units.

This detailed step-by-step calculation shows how to find the radius of a circle when its equation is given in the general form.