Problem 68 - Algebra and Trigonometry

Question

65–72 ? Show that the equation represents a circle, and find the center and radius of the circle. $$ x^{2}+y^{2}+6 y+2=0 $$

Step-by-Step Solution

Answer

The equation is a circle with center (0, -3) and radius \(\sqrt{7}\).

Step 1: Identify the standard form of a circle's equation

The standard form of a circle's equation is given by \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center of the circle and \(r\) is the radius. We need to convert the given equation into this form.

Step 2: Rearrange and group terms

Start by rearranging the given equation: \(x^2 + y^2 + 6y + 2 = 0\). We group \(x^2\) and \(y^2 + 6y\), moving \(2\) to the other side to group constants separately: \(x^2 + y^2 + 6y = -2\).

Step 3: Complete the square for the y terms

Focus on the \(y^2 + 6y\) terms. To complete the square, take the coefficient of \(y\), which is \(6\), halve it to get \(3\), and then square it to get \(9\). Add and subtract \(9\) in the equation: \(x^2 + (y^2 + 6y + 9 - 9) = -2\).

Step 4: Simplify the equation

Complete the square portion: \(x^2 + (y+3)^2 - 9 = -2\). Simplify it further to have \(x^2 + (y+3)^2 = 7\). Now, this equation resembles the standard circle equation with \((y + 3)^2\).

Step 5: Identify the center and radius

From the equation \(x^2 + (y+3)^2 = 7\), we see it is in the form \((x-h)^2 + (y-k)^2 = r^2\). The center \((h, k)\) is \((0, -3)\), and by comparing \(r^2 = 7\), we find \(r = \sqrt{7}\).

Key Concepts

Center of the Circle

In the context of a circle's equation, the center is a crucial concept. By understanding the standard form of a circle's equation, we can identify the center easily. The standard form is \((x-h)^2 + (y-k)^2 = r^2\), where

• \((h, k)\) represents the circle's center,
• and \(r\) is the radius.

To find the center from an equation, we need to rearrange the equation into this form. For example, consider the provided equation: \(x^2 + (y+3)^2 = 7\). Here,

• \(h = 0\),
• and \(k = -3\).

Thus, the center of the circle is at the point \((0, -3)\). By reading off the \((h, k)\) values from the equation, we see how easily the center can be identified once the equation is in the correct form.

Radius of the Circle

Finding the radius of a circle is straightforward once the equation is in standard form. Again, consider the equation \((x-h)^2 + (y-k)^2 = r^2\). Here,

• the term \(r^2\) indicates the square of the radius.

Our example problem simplifies to \(x^2 + (y+3)^2 = 7\). From this,

• we observe that the right side of the equation is \(r^2 = 7\).
• Thus, the radius \(r\) is \(\sqrt{7}\).

It's essential to remember that the radius should always be a positive value, representing the distance from the center to any point on the circle. Therefore, even though the equation gives \(r^2 = 7\), we take the positive square root for the radius. This ensures our understanding aligns with the geometric meaning of a circle's radius.

Completing the Square

The technique of completing the square is fundamental in converting a quadratic equation into standard form, especially for circles. Let's explore how it works: This method helps restructure quadratic expressions so they can fit the form \((x-h)^2\) or \((y-k)^2\). In our exercise, the equation \(x^2 + y^2 + 6y + 2 = 0\) had a tricky part with \(y^2 + 6y\).

• To complete the square, we first took the coefficient of \(y\), which was 6.
• Next, we halved this number, resulting in 3.
• Then we squared this result to get 9.
• We then added and subtracted this squared term (9) inside the equation: \(x^2 + (y^2 + 6y + 9 - 9) = -2\).
• This adjustment allowed us to write it as \((y+3)^2\), which is evident from \(y^2 + 6y + 9 = (y+3)^2\).

Completing the square is not only key for rewriting circles' equations, but it's also valuable for solving various algebraic expressions, making it a versatile tool in mathematics. Understanding this helps bridge the gap between complex expressions and the neat, usable structure of a standard circle equation.