guide:Bef155785d

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We looked at a circle in Chapter 1 and have a definition from a first course in plane geometry. This is still the definition: A circle is the locus of points in a plane at a given distance from a fixed point. The given distance is called the radius and the fixed point is called the center. If the center of the circle is at [math](h, k)[/math], the distance from the center to a variable point [math](x,y)[/math] is, by the distance formula, [math]\sqrt{(x-h)^2 + (y-k)^2}[/math]. If the radius is [math]r[/math], we have an equation of the circle given by

[[math]] \begin{equation} \sqrt{(x-h)^2 + (y-k)^2} = r. \label{eq3.1.1} \end{equation} [[/math]]

An equivalent equation which is more commonly used is

[[math]] \begin{equation} (x-h)^2 + (y-k)^2 = r^2. \label{eq3.1.2} \end{equation} [[/math]]

It is easy to see that, not only do all points at a distance [math]r[/math] from [math](h, k)[/math] lie on the graph of (2), but also all points on the graph of (2) are at a distance r from [math](h, k)[/math].

Example

(a) Write an equation of the circle with center at the origin and radius 3. (b) Write an equation of the circle with center at [math](-1, 2)[/math] and radius 5.

• By the distance formula, the first circle has equation
  [[math]] \sqrt{(x-0)^2 + (y-0)^2} = 3, \;\;\; \mbox{or} \;\;\; x^2 + y^2 = 9. [[/math]]
• By the distance formula, an equation for the second circle is\linebreak [math]\sqrt{[x-(-1)]^2 + (y-2)^2} = 5[/math]. Equivalent equations are
  [[math]] \begin{eqnarray*} (x + 1)^2 + (y - 2)^2 &=& 25, \\ x^2 + y^2 + 2x - 4y &=& 20. \end{eqnarray*} [[/math]]

As the following examples will show, any equation of the form [math]ax^2 + ay^2 + bx + cy + d = 0[/math] is, loosely speaking, an equation of a circle. The words “loosely speaking” are inserted to cover possible degenerate cases. For example, if [math]a = 0[/math] and [math]b[/math] and [math]c[/math] are not both zero, the equation becomes an equation of a line. In another degenerate case “the circle” may be just a point (if its radius is zero), and in another there may be no locus at all.

Example

Describe the graph of each of the following equations:

• [math]x^2 + y^2 - 6x + 8y - 75 = 0[/math],
• [math]x^2 + y^2 + 12x - 2y + 37 = 0[/math],
• [math]x^2+y^2 - 4x- 5y+ 12 = 0[/math],
• [math]3x^2 + 3y^2 - 9x + 10y - \frac{71}{12} = 0[/math].

The technique of completing the square is useful in problems of this type. In (a), we write equations equivalent to the given equation until we recognize the form.

[[math]] \begin{eqnarray*} x^2 - 6x + y^2 + 8y &=& 75, \\ x^2 - 6x + 9 + y^2 + 8y + 16 &=& 75 + 9 + 16, \\ (x - 3)^2 + (y + 4)^2 &=& 100. \end{eqnarray*} [[/math]]

The graph is a circle with center at [math](3, - 4)[/math] and radius 10. Applying the same technique to (b), we have

[[math]] \begin{eqnarray*} x^2 + 12x + y^2 - 2y &=& -37, \\ x^2 + 12x + 36 + y^2 - 2y + 1 &=& -37 + 36 + 1, \\ (x + 6)^2 + (y - 1)^2 &=& 0. \end{eqnarray*} [[/math]]

The graph is a circle with center at [math](- 6, 1)[/math] and radius 0; i.e., it is just the point [math](- 6, 1)[/math]. We may say that the graph consists of the single point, although we sometimes describe it as a point circle. The equation of (c) gives different results:

[[math]] \begin{eqnarray*} x^2 - 4x + y^2 - 5y &=& - 12, \\ x^2 - 4x + 4 + y^2 - 5y + \frac{25}{4} &=& - 12 + 4 + \frac{25}{4}, \\ (x-2)^2 + (y - \frac{5}{2})^2 &=& -\frac{7}{4}. \end{eqnarray*} [[/math]]

For any two real numbers [math]x[/math] and [math]y[/math], the numbers [math](x-2)^2[/math] and [math](y-\frac{5}{2})^2[/math] must both be nonnegative, while [math]-\frac{4}{7}[/math] is certainly negative. Hence there are no points in the plane satisfying this equation. However, the form of the last of the three equivalent equations is that of the equation of a circle, and we sometimes say that the graph is an imaginary circle with center at [math](2, \frac{5}{2})[/math] and radius [math]\frac{1}{2}\sqrt{-7}[/math]. Equation (d) requires a bit more manipulation:

[[math]] \begin{eqnarray*} 3x^2 - 9x + 3y^2 + 10y &=& \frac{71}{12},\\ x^2 - 3x + y^2 + \frac{10}{3}y &=& \frac{71}{36},\\ x^2 - 3x + \frac{9}{4} + y^2 + \frac{10}{3}y + \frac{29}{5} &=& \frac{71}{36} + \frac{9}{4} + \frac{25}{9}, \\ (x - \frac{3}{2})^2 + (y + \frac{5}{3})^2 &=& 7. \end{eqnarray*} [[/math]]

The graph is a circle with center at [math](\frac{3}{2}, -\frac{5}{3})[/math] and radius [math]\sqrt7[/math].

By completing the square, as in the above examples, one can show that any equation of the form

[[math]] ax^2 + ay^2 + bx + cy + d = 0 [[/math]]

is an equation of a circle of positive radius if and only if [math]a \neq 0[/math] and [math]b^2 + c^2 \gt 4ad[/math]. The circle is also the intersection of a right circular cone with a plane perpendicular to the axis of the cone. If the plane passes through the vertex of the cone, the intersection is a point. We can use the techniques of the calculus, as well as our knowledge of Euclidean geometry, to write an equation for a circle or for a line tangent to a circle, from given geometric conditions.

Example

Write an equation of the line which is tangent to the graph of [math](x + 3)^2 + (y - 4)^2 = 25[/math] at (1, 7). We may find the slope by use of the derivative, differentiating implicitly and remembering that one interpretation of the derivative is the slope of the tangent: [math]2(x + 3) + 2(y - 4)y' = 0[/math]; hence [math]y' = - \frac{x + 3}{y - 4}[/math]. The slope Or the tangent is [math]-\frac{1 + 3}{7 - 4} = - \frac{4}{3}[/math]. We may also find the slope of the tangent by noting first that the radius to (1, 7) has slope [math]\frac{7-4}{1- (-3)}= \frac{3}{4}[/math] and then by remembering that the tangent is perpendicular to the radius and hence has slope [math]-\frac{4}{3}[/math]. Thus the tangent line has equation [math]y-7 = - \frac{4}{3}(x-1)[/math] or [math]4x + 3y = 25[/math].

General references

Doyle, Peter G. (2008). "Crowell and Slesnick's Calculus with Analytic Geometry" (PDF). Retrieved Oct 29, 2024.