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 of the circle is (0, 3) and the radius is 5.

Step 1: Rewrite the Circle Equation

The given equation is \(x^{2} + y^{2} - 6y = 16\). It doesn't immediately reveal the center and the radius of the circle. To find these, we need to rewrite it in the standard form of a circle equation: \((x-h)^{2} + (y-k)^{2} = r^{2}\).

Step 2: Complete the Square for y

Focus on the \(y\) terms. Isolate them: \(y^{2} - 6y\). To complete the square, take half of the coefficient of \(y\), which is \(-6\), divide by 2 to get \(-3\), and then square it to get \(9\). Add and subtract \(9\) inside the equation.This modifies our equation to: \(x^{2} + y^{2} - 6y + 9 - 9 = 16\).

Step 3: Simplify the Equation

By completing the square, our equation becomes:\[x^{2} + (y-3)^{2} - 9 = 16\]Add \(9\) to both sides to get:\[x^{2} + (y-3)^{2} = 25\]

Step 4: Identify the Center and Radius

The equation \(x^{2} + (y-3)^{2} = 25\) matches the standard form of a circle \((x-h)^{2} + (y-k)^{2} = r^{2}\). Here, \(h = 0\) and \(k = 3\), so the center \((h,k)\) is \((0,3)\). The radius \(r\) is the square root of 25, which is 5.

Key Concepts

Completing the Square

Completing the square is a technique used in algebra to change the way a quadratic expression is written. This can help reveal certain properties of the expression, like its minimum or maximum value. It's particularly useful to rewrite a circle equation in its standard form. Here's how it works step-by-step:

• Start with the terms you want to complete, like in our exercise: \(y^2 - 6y\).
• Take the coefficient of the linear term, half it, then square it. So, for \(-6y\), half of \(-6\) is \(-3\), and squaring \(-3\) gives you \(9\).
• Incorporate this square into the equation by adding and subtracting it within the expression, ensuring the balance of the equation isn't affected. Here, you'd add and subtract \(9\) right next to \(y^2 - 6y\), resulting in \((y^2 - 6y + 9) - 9\).

By completing the square on a circle's equation, you can transform it into a form that gives direct insights into the geometry of the circle.

Standard Form of a Circle

The standard form of a circle's equation is a simple and elegant way to represent all the important aspects of a circle using algebra. It is expressed as: \[(x-h)^2 + (y-k)^2 = r^2\]Here's a breakdown of what each part means:

• \((h, k)\) represents the center of the circle. This is where the circle is centered on the coordinate plane.
• \(r\) is the radius of the circle. It tells you how far any point on the circle is from its center.

In our example equation, we rewrote \[x^{2} + (y-3)^{2} = 25\]to match this standard form. This rewriting process allows us to easily identify both the center and the radius of the circle. So, once the equation is converted to this form, determining these properties becomes straightforward.

Finding Center and Radius

Identifying the center and radius of a circle from its equation is easy once the equation is in the standard form. The term \((x-h)^2 + (y-k)^2 = r^2\) directly points out the circle's center and its radius.

When you look at the equation: \[x^{2} + (y-3)^{2} = 25\]

• The center \((h, k)\) can be found by identifying the values subtracted from \(x\) and \(y\). In this case, there's no \(x\) term being added or subtracted, meaning \(h = 0\); and \(y\) has a \(-3\), giving us \(k = 3\).
• The radius \(r\) is the square root of the number on the right side of the equation. So, with \(25\) on the right side, \(r = \sqrt{25} = 5\).

By following this process for any circle equation, you can efficiently determine the center and radius, helping visualize and work with the shape on the plane clearly.