Problem 13 - The Calculus with Analytic Geometry

Question

In Exercises 11 through 14 , find the center and radius of each circle, and draw a sketch of the graph. $$ 3 x^{2}+3 y^{2}+4 y-7=0 $$

Step-by-Step Solution

Answer

Center: (0, -2/3); Radius: 5/3

Step 1: Rewrite the equation

First, rewrite the given equation to group the x and y terms together. Divide the entire equation by 3 to simplify: \[ x^2 + y^2 + \frac{4}{3}y - \frac{7}{3} = 0 \]

Step 2: Complete the square for the y terms

To complete the square for the y-terms, isolate the constant on one side: \[ x^2 + y^2 + \frac{4}{3}y = \frac{7}{3} \] Add and subtract the square of half the coefficient of y inside the equation: \[ y^2 + \frac{4}{3}y + \left(\frac{2}{3}\right)^2 = \frac{7}{3} + \left(\frac{2}{3}\right)^2 \] Thus, it becomes: \[ y^2 + \frac{4}{3}y + \frac{4}{9} = \frac{21}{9} + \frac{4}{9} \] Which simplifies to: \[ y^2 + \frac{4}{3}y + \frac{4}{9} = \frac{25}{9} \]

Step 3: Write the equation in the standard form of a circle

Now express the completed square as a binomial square: \[ x^2 + \left( y + \frac{2}{3} \right)^2 = \frac{25}{9} \] The standard form of a circle equation is \( (x-h)^2 + (y-k)^2 = r^2 \). Here, \( h = 0 \), \( k = -\frac{2}{3} \), and \( r = \frac{5}{3} \). Hence, the equation of the circle is: \[ (x-0)^2 + \left( y + \frac{2}{3} \right)^2 = \left( \frac{5}{3} \right)^2 \]

Step 4: Identify the center and radius

From the standard form, the center \((h,k)\) of the circle is \( (0, -\frac{2}{3}) \) and the radius \( r \) is \( \frac{5}{3} \).

Step 5: Draw the graph

Sketch the circle with the identified center at \( (0, -\frac{2}{3}) \) and the radius of \( \frac{5}{3} \). Draw a circle starting from the center and extending \( \frac{5}{3} \) units away in all directions.

Key Concepts

Completing the Square

Completing the square is a method used to transform a quadratic equation into a perfect square trinomial. This is helpful in finding the standard form of a circle's equation. For example, in the given problem:
Step 1 involves isolating the y-terms and completing the square for them. To complete the square, you:

• Take the coefficient of the linear term (in this case, \frac{4}{3}\) and halve it.
• Square that result (\(\frac{2}{3}\) becomes \(\frac{4}{9}\)).
• Add and subtract this square inside the equation to maintain equality.

This allows the quadratic expression \(y^2 + \frac{4}{3} y + \frac{4}{9}\) to be written as a binomial square \( (y + \frac{2}{3})^2 \, making it easier to identify the center and radius of the circle.

Graphing Circles

Graphing a circle involves plotting all points that are at a fixed distance (the radius) from a central point (the center). From the completed square equation \( x^2 + ( y + \frac{2}{3})^2 = \frac{25}{9} \), we know that:

• The center is at \(0, -\frac{2}{3}\)
• The radius is \( \frac{5}{3} \)

When you graph this, remember:

• Locate the center on the coordinate plane.
• Measure the radius from the center in all directions to mark the circumference.
• Draw a smooth curve through these points to create the circle.

Sketching the graph helps visually verify the solution and understand the geometry involved.

Radius and Center of a Circle

The radius and center of a circle are key components in its equation. Given the standard form:\( (x-h)^2 + (y-k)^2 = r^2 \),

• The center \(h, k\) represents the coordinates from which every point on the circle is equidistant.
• The radius \(r\) is the length of that fixed distance.

In our problem, we identified the standard form as \( (x-0)^2 + (y+\frac{2}{3})^2 = (\frac{5}{3})^2 \), Revealing:

• The center \((h, k)\) is at \( (0, -\frac{2}{3}) \).
• The radius \((r)\) is \( \frac{5}{3} \).

Understanding these components is crucial for graphing and solving circle-related problems. Always check the structure of the equation to correctly identify these elements.