Problem 79 - College Algebra

Question

How is the standard form of a circle's equation obtained from its general form?

Step-by-Step Solution

Answer

The standard form of a circle's equation is obtained from its general form by grouping and completing the square of the x and y terms to reduce it to a format of \( (x-h)^2 + (y-k)^2 = r^2 \), which represents a circle with centre (h,k) and radius r.

Step 1: Identify the General Form of a Circle

The general form of a circle is given as \(Ax^2 + By^2 + Cx + Dy + E = 0\), where A and B are equal and C, D, and E are constants. For simplicity, we take A=B=1 and the equation takes the form of \(x^2 + y^2 + Cx + Dy + E = 0\).

Step 2: Re-arrange the equation

Rearrange the equation to group the x and y terms together: \(x^2 + Cx + y^2 + Dy + E = 0\).

Step 3: Get into the standard form

In the standard form equation of a circle, \( (x-h)^2 + (y-k)^2 = r^2 \), 'h' and 'k' are the coordinates of the center and 'r' is the radius of the circle. Complete the square on the x and y terms to convert the equation into the standard form. For the x-terms the square completion is done by adding and subtracting \((C/2)^2\) inside the x-terms and for y-terms \((D/2)^2\). This results in the equation: \((x + C/2)^2 - (C/2)^2 + (y + D/2)^2 - (D/2)^2 + E = 0\)

Step 4: Complete the conversion

Re-arrange the equation in the form \( (x-h)^2 + (y-k)^2 = r^2 \) to get the standard form: \((x + C/2)^2 + (y + D/2)^2 = (C/2)^2 + (D/2)^2 -E \). Therefore, \( h = -C/2\), \( k = -D/2 \) and radius \( r = sqrt[((C/2)^2 + (D/2)^2 - E)] \).