Problem 79 - College Algebra 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, which is \((x-h)^2 + (y-k)^2 = r^2\), is obtained from its general form, \(Ax^2 + By^2 + Cx + Dy + E = 0\), by identifying the coefficients, completing the square and consequently identifying the center \((-C/2, -D/2)\) and the radius \(\sqrt{(C/2)^2 + (D/2)^2 - E}\).

Step 1: Identify the Coefficients

Firstly, the coefficients of the general form should be well-identified. The equation is given by \(Ax^2 + By^2 + Cx + Dy + E = 0\). Here, A, B, C, D, and E are the coefficients to identify.

Step 2: Convert to Standard Form

Assuming that the coefficients A and B are 1 (since in a circle equation, they must be equal, thus for simplification we take them as 1), the equation can be rewritten as \(x^2 + y^2 + Cx + Dy + E = 0\). The next step to convert this to the standard form \((x-h)^2 + (y-k)^2 = r^2\) is by completing the square. Firstly, group the x terms and the y terms together, giving \(x^2 + Cx + y^2 + Dy + E = 0\). Next, rewrite the equation by completing the square which gives \((x + C/2)^2 - (C/2)^2 + (y + D/2)^2 - (D/2)^2 = -E\). After simplifying, the equation in standard form becomes \((x + C/2)^2 + (y + D/2)^2 = (C/2)^2 + (D/2)^2 - E\).

Step 3: Identify the Center and the Radius

Finally, identify the center and the radius of the circle. Comparing the standard form equation \((x-h)^2 + (y-k)^2 = r^2\) to \((x + C/2)^2 + (y + D/2)^2 = (C/2)^2 + (D/2)^2 - E\), the centre of the circle is identified at \((-C/2, -D/2)\) and the radius is \(\sqrt{(C/2)^2 + (D/2)^2 - E}\).