Problem 79 - Precalculus Essentials

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, \( (x - h)^2 + (y - k)^2 = r^2 \) is obtained from its general form, \(Ax^2 + By^2 + Cx + Dy + E = 0\), by completing the square. Here, h = -C/2A, k = -D/2B, and \( r = \sqrt{(C/2A)^2+(D/2B)^2 - E} \).

Step 1: General Form of a Circle Equation

The general form of a circle's equation is \(Ax^2 + By^2 + Cx + Dy + E = 0\), where A equals B, and A and B are both not equal to zero.

Step 2: Completing the Square

We need to complete the square in this scenario. Group x's and y's together, the general form becomes \((A)x^2 + Cx + (B)y^2 + Dy + E = 0\). After that, rewrite it as \((A)(x^2 + (C/A)x) + (B)(y^2 + (D/B)y) = -E)\.

Step 3: Transform to Standard Form

You can complete the square in the x's and y's terms by adding \( (C/2A)^2 \) for x's and \( (D/2B)^2 \) for y's on both sides of the equation respectively. Thus, the equation transforms into its standard form \((x - h)^2 + (y - k)^2 = r^2\), where h = -C/2A, k = -D/2B, and \( r = \sqrt{(C/2A)^2+(D/2B)^2 - E} \).