Problem 18 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 17-22, find the center and radius of the circle with the given equation. x^{2}+y^{2}-6 y=16

Step-by-Step Solution

Answer

The center is (0, 3) and the radius is 5.

Step 1: Identify the Standard Equation of the Circle

The standard equation of a circle is \( (x-h)^2 + (y-k)^2 = r^2 \), where \((h, k)\) represents the center of the circle and \(r\) is the radius. Our goal is to transform the given equation into this form.

Step 2: Complete the Square for the y-Term

The given equation is \( x^2 + y^2 - 6y = 16 \). We leave \(x^2\) as is because completing the square will be applied to terms involving \(y\). First, we rewrite \( y^2 - 6y \) such that we can complete the square as follows. To complete the square, find the number \( (\text{-Coefficient of } y / 2)^2 = (-6/2)^2 = 3^2 = 9 \). Add and subtract 9 inside the equation: \( x^2 + (y^2 - 6y + 9 - 9) = 16 \). This modifies the equation to \( x^2 + (y - 3)^2 - 9 = 16 \).

Step 3: Simplify and Arrange

Add 9 to both sides of the equation: \( x^2 + (y - 3)^2 = 16 + 9 \). Thus, the equation becomes \( x^2 + (y - 3)^2 = 25 \).

Step 4: Identify the Center and Radius

Now the equation is in the standard form \( (x-h)^2 + (y-k)^2 = r^2 \), where \( h = 0 \) and \( k = 3 \), resulting in the center being \( (0, 3) \). The right side of the equation, 25, represents \( r^2 \), so the radius \( r \) is \( \sqrt{25} = 5 \).

Key Concepts

Equation of a Circle

An equation of a circle is a straightforward mathematical expression used to describe all the points that form a circle in a two-dimensional plane. The standard equation of a circle is given by: \[(x-h)^2 + (y-k)^2 = r^2\]This equation is composed of specific parts:

• \((h, k)\) - The coordinates of the circle's center.
• \(r\) - The radius of the circle, which is the distance from the center to any point on the circle.

The equation clearly states that if you take any point \((x, y)\) on the circle, the distance from that point to the center \((h, k)\) should equal the radius \(r\). This forms a perfect circular shape when graphed.
To transform any quadratic equation into this standard form, you often need to use a method called completing the square. Completing the square helps in grouping terms to form a perfect square, which is instrumental in identifying the center and radius of the circle.

Center of a Circle

The center of a circle, represented by the coordinates \((h, k)\) in the standard circle equation, is crucial for understanding its position in the coordinate plane. By examining the equation \[(x-h)^2 + (y-k)^2 = r^2\]the center \((h, k)\) serves as the starting point from which every point on the circle is equidistant. The position of the center dictates where the circle is located within the graph.
For example, if the equation is brought to the form \((x-0)^2 + (y-3)^2 = r^2\), the center of the circle becomes apparent as \((0, 3)\). This shows that the circle is centered right on the y-axis, three units above the x-axis origin.
Knowing the center helps students quickly graph the circle by marking its focal point and then plotting points, ensuring the same distance (the radius) from this center, to form the circle's outline.

Radius of a Circle

The radius of a circle, denoted by \(r\) in the circle's standard equation, is the length from the center to any point on the circle. It is a constant value that defines the size of the circle and is always non-negative. According to the equation:\[(x-h)^2 + (y-k)^2 = r^2\]the right side \(r^2\) is crucial in calculating the radius. Understanding that this portion equals the square of the radius allows us to derive the radius itself by taking the square root.
In our specific problem, after completing the square and simplifying, we find that:\[(x-0)^2 + (y-3)^2 = 25\]Here, \(r^2 = 25\), making the radius \(r = \sqrt{25} = 5\).The radius gives clear information about how "big" the circle is. By knowing this size, one can not only sketch the circle accordingly but also better understand spatial relationships in related mathematical problems or real-world applications.