Problem 60 - College Algebra

Question

Complete the square and write the equation in standard form. Then give the center and radius of each circle and graph the equation. $$ x^{2}+y^{2}-6 y-7=0 $$

Step-by-Step Solution

Answer

The equation in standard form is \(x^{2}+(y-3)^2=16\). Hence, the center of the circle is at point (0, 3) and has a radius of 4.

Step 1: Rewrite the Equation

Firstly, the equation needs to be rewritten by grouping the y-terms together \(x^{2}+(y^{2}-6y)-7=0\). Then, you can complete the square.

Step 2: Complete the Square

To complete the square for \(y^2-6y\), take half the coefficient of y, square it, and add it to both sides of the equation. Half of -6 is -3, and (-3)^2 = 9 so the equation becomes: \(x^{2}+(y^{2}-6y+9)-7=9\). This simplifies to \(x^{2}+(y-3)^2=16\).

Step 3: Determine the Center and Radius

In standard form, the general equation of a circle is \( (x-h)^2 + (y-k)^2 = r^2 \) where the center is given by (h,k) and r is the radius. By comparing this generalized form with our simplification we get the center (h,k) as (0,3) and radius 'r' as \(\sqrt{16}\) = 4.

Step 4: Graph the Equation

When you plot (0,3) as the center of the circle and use 4 as its radius, the graph of the equation will be a circle centered at (0,3) and with a radius of 4 units.

Key Concepts

Standard Form of a Circle

When it comes to understanding the structure of a circle's equation, the standard form is an essential concept. In its simplest terms, the standard form equation for a circle is \( (x - h)^2 + (y - k)^2 = r^2 \), where \( h \) and \( k \) are the coordinates of the circle's center, and \( r \) is the radius. The beauty of the standard form is that it neatly encapsulates all of the crucial circle properties in a single equation.

To illustrate how this works in practice, let's consider a typical example. If you start with an equation like \( x^2 + y^2 - 6y - 7 = 0 \), it's not immediately clear what the center or the radius of the circle is. By completing the square, we can rearrange the equation into the standard form. Completing the square transforms the equation into a neat package that clearly displays the relevant attributes of the circle. This makes it much easier to analyze and solve problems related to circles.

Center and Radius of a Circle

The center and radius are two fundamental characteristics of a circle. The center is a fixed point from which every point on the circle is equidistant. In the standard form equation, the center's coordinates are represented by \( (h, k) \). The radius is the constant distance from the center to any point on the circle, denoted by \( r \) in the equation.

For the given exercise, after completing the square to reach the form \(x^{2}+(y-3)^2=16\), we immediately know the center is at (0, 3). This comes from the understanding that \(h = 0\) and \(k = 3\), as there is no \(x\)-term affecting the center’s \(h\)-coordinate. The radius is found by taking the square root of \(16\), giving us a radius of 4 units. With this approach to identifying the center and radius from the standard form, you can easily chart out the basic shape and location of any circle on a coordinate plane.

Graphing Equations

Graphing equations is a visual way to represent mathematical concepts. For circles, once you have the standard form of the equation and know the center and radius, graphing is straightforward. Start by plotting the center on the coordinate plane. Then, using the radius, draw a circle that extends that distance from the center in all directions.

In our scenario, with the center at (0, 3) and a radius of 4, you would begin by marking the point (0, 3) on the graph. This is your circle's midpoint. From there, move 4 units up, down, left, and right, marking these points. Connect these points in a round shape to form your circle. Remember, every point on this curve is 4 units away from the center, adhering to our radius. Graphing helps bring the equation to life and solidifies understanding of the circle's size and position in space.