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Chapter 13: Problem 41

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

Short Answer

Expert verified
The standard form of a circle's equation, \((x - h)^2 + (y - k)^2 = r^2\), is obtained from its general form, \(x^2 + y^2 + cx + dy + e = 0\), by rearranging terms and completing the square for the \(x\) and \(y\) terms.

Step by step solution

01

General Form

The general form of a circle's equation is: \(ax^2 + by^2 + cx + dy + e = 0\). However, for a circle, \(a\) is equal to \(b\) and both are equal to 1. Therefore, the general form for a circle specifically becomes: \(x^2 + y^2 + cx + dy + e = 0\).
02

Rearranging terms

The terms in \(x\) and \(y\) are organized together, which gives the equation as: \(x^2 + cx + y^2 + dy = -e\). We are aiming to express this equation in the form \((x-h)^2 + (y-k)^2 = r^2\), where \((h,k)\) are the coordinates of the center and \(r\) is the radius of the circle.
03

Completing the square

To make the expression \(x^2 + cx\) a perfect square, we add the square of half of \(c\), i.e., \((c/2)^2\), to both sides of the equation. Similarly, for \(y^2 + dy\), add the square of half of \(d\), i.e., \((d/2)^2\), to both sides. The equation becomes: \(x^2 + cx + (c/2)^2 + y^2 + dy + (d/2)^2 = -e + (c/2)^2 + (d/2)^2\).
04

Final standard form

Now, the equation can be rewritten using \((a+b)^2 = a^2 + 2ab + b^2\) as: \((x + c/2)^2 + (y + d/2)^2 = -e + (c/2)^2 + (d/2)^2\). The equation is now in standard form, with \((h, k) = (-c/2, -d/2)\) and \(r = \sqrt{-e + (c/2)^2 + (d/2)^2}\).

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