Problem 51 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 47-56: Find the center and radius of the circle with the given equation. $$ 2 x^{2}+2 y^{2}-12 x+4 y-15=0 $$

Step-by-Step Solution

Answer

The center of the circle is (3, -1) and the radius is \(\frac{3\sqrt{6}}{2}\).

Step 1: Rewrite the Equation in Standard Form

First, divide every term in the equation by 2 to simplify it. The equation becomes:\[x^2 + y^2 - 6x + 2y - \frac{15}{2} = 0\]

Step 2: Move Constant to the Other Side

Move the constant term to the right side of the equation:\[x^2 - 6x + y^2 + 2y = \frac{15}{2}\]

Step 3: Complete the Square for x-terms

To complete the square for the \(x\)-terms, take half of the \(-6\) coefficient, square it, and add it to both sides:\[x^2 - 6x \quad \Rightarrow \quad (x - 3)^2 = x^2 - 6x + 9 \ \]Add 9 to both sides of the equation.

Step 4: Complete the Square for y-terms

Similarly, for the \(y\)-terms, take half of the \(2\) coefficient, square it, and add it to both sides:\[y^2 + 2y \quad \Rightarrow \quad (y + 1)^2 = y^2 + 2y + 1 \ \]Add 1 to both sides of the equation.

Step 5: Write the Equation in Completed Square Form

After completing the square, the equation is:\[(x - 3)^2 + (y + 1)^2 = \frac{15}{2} + 9 + 1 = \frac{27}{2}\]Thus, the equation is in the form \((x - h)^2 + (y - k)^2 = r^2\).

Step 6: Identify the Center and Radius

From the standard form equation \((x - h)^2 + (y - k)^2 = r^2\), we identify:- Center \((h, k) = (3, -1)\)- Radius \(r = \sqrt{\frac{27}{2}}\)

Step 7: Simplify the Radius Result if Possible

Simplify the expression for the radius:\[ r = \sqrt{\frac{27}{2}} = \sqrt{\frac{54}{4}} = \frac{\sqrt{54}}{2} = \frac{3\sqrt{6}}{2}\]Thus, the radius is \(\frac{3\sqrt{6}}{2}\).

Key Concepts

Completing the Square

Completing the square is a technique used to transform a quadratic expression into a perfect square trinomial. This is particularly useful in geometry when dealing with circle equations, as it helps express the equation in a form where you can easily identify the circle's center and radius.
To complete the square, follow these steps:

• Identify the coefficient of the linear term in the expression.
• Take half of this coefficient, and then square it.
• Add and subtract this squared value within the expression.

For example, when dealing with the term \(x^2 - 6x\), take half of \(-6\), which is \(-3\), and then square it to get \(9\). Add \(9\) to complete the square, which turns the expression into \((x - 3)^2\).
This process not only simplifies equations but repositions terms to reveal critical characteristics of conic shapes, like circles.

Standard Form of Circle Equation

The standard form of a circle's equation is given by: \((x - h)^2 + (y - k)^2 = r^2\). This form draws directly from completing the square, allowing us to view the circle's geometrical properties simply.
In this equation:

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

By transforming a complex equation into this form, you not only simplify your work but also unlock the potential to directly interpret and graph the circle. For instance, after completing the square for both the \(x\) and \(y\) terms, you can rearrange them into this standard form. This helps to easily identify identifying the circle's essential characteristics from its equation.

Center and Radius of Circle

Once the equation of a circle is in the standard form, identifying the center and radius becomes straightforward.

• The center \((h, k)\) of the circle is found from the terms \((x - h)\) and \((y - k)\).
• The radius \(r\) is determined from the right side of the equation \(r^2\).

Understanding the circle's center and radius is critical in both algebra and geometry because:
It helps in graphing the circle accurately; locating the center \((3, -1)\) in our example, and measuring out the distance of the radius ensures precision in sketching.
Solving problems involving circular shapes becomes easy, as you can rely on these coordinates and measures to navigate through multiple geometry tasks involving circles.