Problem 54 - Calculus Early Transcendentals: Pearson New International Edition

Question

What relationship between \(a, b\), and \(c\) must hold if \(x^{2}+a x+y^{2}+b y+c=0\) is the equation of a circle?

Step-by-Step Solution

Answer

\( c = \frac{a^2}{4} + \frac{b^2}{4} - r^2 \) with \( r^2 > 0 \).

Step 1: Identify the General Circle Equation

The general equation of a circle in the form of \( (x - h)^2 + (y - k)^2 = r^2 \) can be expanded to give \( x^2 + y^2 - 2hx - 2ky + h^2 + k^2 = r^2 \). By rewriting it to match the given equation, we have \( x^2 + 2(-h)x + y^2 + 2(-k)y + (h^2 + k^2 - r^2) = 0 \). Here, \( h \), \( k \) are the circle’s center coordinates, and \( r \) is the radius.

Step 2: Match Coefficients

Compare the coefficients of similar terms between the expanded circle equation and the given equation \( x^2 + ax + y^2 + by + c = 0 \). This comparison yields: \( a = -2h \), \( b = -2k \), and \( c = h^2 + k^2 - r^2 \).

Step 3: Express Conditions in Terms of Coefficients

Express \( h \) and \( k \) in terms of \( a \) and \( b \): \( h = -\frac{a}{2} \) and \( k = -\frac{b}{2} \). Substituting these values into the equation for \( c \), we get: \( c = \left(-\frac{a}{2}\right)^2 + \left(-\frac{b}{2}\right)^2 - r^2 = \frac{a^2}{4} + \frac{b^2}{4} - r^2 \).

Step 4: Formulate the Final Relationship

The requirement for the equation to be that of a circle is \( c = \frac{a^2}{4} + \frac{b^2}{4} - r^2 \). Rearranged, it is necessary for \( c = \frac{a^2}{4} + \frac{b^2}{4} - r^2 \). For an actual circle, \( r^2 \) must be positive.

Key Concepts

General Circle Equation

Understanding the equation of a circle is essential when dealing with circles in mathematics. The general equation of a circle in its expanded form is \[(x - h)^2 + (y - k)^2 = r^2\] where \(h\) and \(k\) are the coordinates of the center of the circle, and \(r\) is the radius of the circle. This equation represents all the points \((x, y)\) that are at a distance \(r\) from the center.
If we expand this equation, we get:\[x^2 - 2hx + h^2 + y^2 - 2ky + k^2 = r^2\]. From there, we can match this with the standard form of second-order polynomials to relate to the problem provided.

Coefficients Comparison

Comparing coefficients is a crucial step in transforming equations. For the given equation \(x^2 + ax + y^2 + by + c = 0\), the goal is to identify how its coefficients relate to the ones in the expanded form of the circle equation.
By aligning coefficients, you can find:

• \(a = -2h\)
• \(b = -2k\)
• \(c = h^2 + k^2 - r^2\)

These relationships give us the connection between the linear terms, \(x\) and \(y\), and the circle's geometrical parameters such as center and radius.
Thus, coefficient comparison acts as a bridge connecting algebraic manipulation to geometric interpretation.

Circle Center and Radius

Determining the center and radius of a circle from its equation is a common task in geometry. Using the relationships derived from the coefficients:

• \(h = -\frac{a}{2}\)
• \(k = -\frac{b}{2}\)

This shows that the center of the circle \((h, k)\) can be easily determined by halving and negating the coefficients \(a\) and \(b\).
For the radius, rearrange the equation for \(c\): \[r^2 = \frac{a^2}{4} + \frac{b^2}{4} - c\] Thus, \(r\) is derived from the sum of squares of \(a/2\) and \(b/2\) minus \(c\), ensuring that \(r^2\) is positive to confirm a real circle exists. This balance confirms the unique geometrical properties embodied in the circle's algebraic equation.