Problem 60 - Precalculus Essentials

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 circle's center is at (0,3) and the radius is 4 units. The graph should be a circle centered at (0,3) with radius 4 units.

Step 1: Arrange the equation

First arrange the equation \(x^{2}+y^{2}-6y-7=0\) so that the y-squared term and y are together: \(x^{2} + (y^{2}-6y) - 7 = 0\) .

Step 2: Complete the Square

Complete the square for the y terms. The coefficient of y is -6, so (-6/2)^2 = 9. Add this inside the bracket to complete the square: \(x^{2} + (y^{2}-6y+9) - 7 = 9\). This gives: \(x^{2} + (y-3)^{2} = 16 \).

Step 3: Identify Center and Radius

In this standard form, the center of the circle is at the point (h, k) and the radius is r. By comparing with the standard form, we get the center as (0, 3) and radius as 4 (since r = \( \sqrt{16} \) ).

Step 4: Graphing the Equation

Plot the center at (0,3). Draw a circle with radius 4, centered at (0, 3). Plot points at a distance of 4 units in all directions from the center to find the circle.

Key Concepts

Equation of a Circle

The equation of a circle reveals key information about its properties. Typically, it's expressed using variables that represent the position (center) and size (radius) of the circle. The standard mathematical form for a circle’s equation is based on these two primary features.
By understanding this equation, you can identify where the circle is located in coordinate space and its extent.
For example, in the equation \(x^2 + y^2 - 6y - 7 = 0\), you rearrange and complete the square for one of the variables (either x or y) to write it in a more recognizable form.
This arrangement allows you to then extract vital details about the circle itself, like its center and radius, which are fundamental in drawing and understanding the circle’s orientation in space.

Standard Form

The standard form of a circle’s equation helps simplify the process of graphing and interpreting the circle's properties. The standard form is: \[ (x - h)^2 + (y - k)^2 = r^2 \] In this equation:

• \((h, k)\) - represents the center of the circle.
• \(r\) - represents the radius of the circle.

By completing the square, you transform a general equation into this standard form.
In our example, \((y^2 - 6y)\) was completed to become \((y - 3)^2\). This moves the expression to \(x^2 + (y - 3)^2 = 16\), illustrating a circle centered at \((0, 3)\) with a radius squared of 16. This final form makes it much easier to both visualize and plot the circle on a graph.

Center and Radius of a Circle

Knowing the center and radius of a circle gives you a complete description of the circle’s properties. The center, \((h, k)\), is the exact middle point of the circle, while the radius, \(r\), is the constant distance from the center to any point on the circle.
In the example, the completed equation \(x^2 + (y - 3)^2 = 16\) shows that the center is \((0, 3)\) and the radius is 4, derived from \(\sqrt{16}\).
Determining these values allows you to accurately plot the circle. You start at the center and use the radius to establish the circle's boundary, ensuring it retains a perfect round shape.
These components are fundamental to understanding circles in geometry and help apply the concept in real-world problems and mathematical contexts.