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

Step 1: Identify the General Form of the Circle Equation

The given equation is \(x^2 + y^2 + 6y + 2 = 0\). The general form of a circle's equation is \(x^2 + y^2 + Dx + Ey + F = 0\). The given equation lacks an \(x\) term, indicating no horizontal shift in the circle.

Step 2: Rearrange and Prepare to Complete the Square

To complete the square for the \(y\) terms, first isolate them: \(y^2 + 6y\). To begin completing the square, you need to take half of the coefficient of \(y\), which is 6, then square it.

Step 3: Complete the Square for the y-Term

Take the coefficient of \(y\), which is 6, divide by 2 to get 3, and square it to get 9. Rewrite the \(y\) terms as \((y+3)^2 - 9\). This gives us: \(x^2 + (y+3)^2 - 9 + 2 = 0\).

Step 4: Simplify the Equation

Substitute \((y+3)^2 - 9\) back into the equation: \(x^2 + (y+3)^2 - 9 + 2 = 0\). Simplify it to \(x^2 + (y+3)^2 = 7\).

Step 5: Identify the Center and Radius of the Circle

The standard form of a circle's equation is \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius. Comparing with \(x^2 + (y+3)^2 = 7\), the center is \((0, -3)\) and the radius \(r\) is \(\sqrt{7}\).

Key Concepts

Completing the Square

When turning a general quadratic equation into the equation of a circle, completing the square is a useful technique. This method helps us rewrite a quadratic expression in a form that reveals different features, like the center and radius of a circle. Here's how it works with the equation given:

• Start by isolating the terms that need to be squared. In the equation \(x^2 + y^2 + 6y + 2 = 0\), we focus on \(y^2 + 6y\).
• Take the coefficient of \(y\), which is 6, divide it by 2 to get 3. Then, square 3 to obtain 9.
• Rewrite \(y^2 + 6y\) as \((y + 3)^2 - 9\) to maintain the equivalence.

After completing the square, the equation transforms into \(x^2 + (y+3)^2 - 9 + 2 = 0\). Simplifying, we end up with \(x^2 + (y+3)^2 = 7\). This form helps us see the structure of a circle equation.

Center of a Circle

The center of a circle is a key feature that can be easily identified once the equation is in the standard form. The standard format for a circle's equation is \((x-h)^2 + (y-k)^2 = r^2\). Here, \((h, k)\) represents the circle's center.

• From the equation \(x^2 + (y+3)^2 = 7\), compare it with the standard form.
• The term \((y+3)^2\) indicates a transformation in the \(y\)-direction. Specifically, it shifts the center to \(-3\) in the \(y\)-axis.
• Notice that there is no \(x\) transformation like \((x-h)^2\). Therefore, \(h = 0\), meaning there is no shift horizontally.

Thus, the center of the circle in the equation is \((0, -3)\). Understanding this helps in visualizing the circle's position in a coordinate plane.

Radius of a Circle

The radius of a circle is the distance from the center to any point on the circle. It can be directly derived from the standard circle equation \((x-h)^2 + (y-k)^2 = r^2\), where \(r\) is the radius.

• Look again at the rewritten equation: \(x^2 + (y+3)^2 = 7\).
• In this form, \(r^2\) is equivalent to 7. To find the actual radius, take the square root of 7.
• Thus, the radius \(r\) is \(\sqrt{7}\).

Comprehending the radius allows you to determine the size of the circle. It's an essential component that measures how far the edge of the circle extends from its center. This understanding is crucial for solving many geometry-related problems.