Problem 70 - Algebra and Trigonometry

Question

65–72 ? Show that the equation represents a circle, and find the center and radius of the circle. $$ x^{2}+y^{2}+2 x+y+1=0 $$

Step-by-Step Solution

Answer

The circle's center is \((-1, -\frac{1}{2})\), and its radius is \(\frac{\sqrt{3}}{2}\).

Step 1: Recognize the Standard Form of a Circle

The standard form equation of a circle is \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius of the circle.

Step 2: Complete the Square for x-terms

Given the equation \(x^2 + y^2 + 2x + y + 1 = 0\), start by grouping the \(x\)-terms: \(x^2 + 2x\). To complete the square for \(x^2 + 2x\), add and subtract \((\frac{2}{2})^2 = 1\):\[(x^2 + 2x + 1) - 1 = (x+1)^2 - 1.\]

Step 3: Complete the Square for y-terms

Now, group the y-terms: \(y^2 + y\). To complete the square, add and subtract \((\frac{1}{2})^2 = \frac{1}{4}\):\[(y^2 + y + \frac{1}{4}) - \frac{1}{4} = (y+\frac{1}{2})^2 - \frac{1}{4}.\]

Step 4: Rewrite the Equation with Completed Squares

Now rewrite the entire equation using completed squares: \((x+1)^2 - 1 + (y+\frac{1}{2})^2 - \frac{1}{4} + 1 = 0.\) Simplify by combining constants on the right side of the equation:\[(x+1)^2 + (y+\frac{1}{2})^2 = 1 - \frac{1}{4} - 1 = \frac{3}{4}.\]

Step 5: Identify the Center and Radius

The equation \((x+1)^2 + (y+\frac{1}{2})^2 = \frac{3}{4}\) is now in standard form. Hence, the center of the circle is \((-1, -\frac{1}{2})\), and the radius is \(\sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}.\)

Key Concepts

Completing the Square

Completing the square is a key algebraic technique for rewriting a quadratic equation. It helps in identifying certain patterns and can turn a complex equation into an easier-to-solve form.
When we are given a quadratic equation in the form of

• \( x^2 + bx \) or
• \( y^2 + cy \)

we complete the square by adding and subtracting a specific value. This value is

• \(( \frac{b}{2} )^2\) for the x-terms, and
• \(( \frac{c}{2} )^2\) for the y-terms.

Adding this value transforms part of the quadratic expression, such as \(x^2 + bx\), into \((x + \frac{b}{2})^2\). By maintaining the equation's balance, we either add or subtract this same value on both sides.
This simplification is crucial when working with conics, such as circles, because it reveals the hidden structure within the equation.

Standard Form of a Circle

The standard form of a circle's equation is a neat representation that makes it easy to locate the circle's center and radius. The general form is \((x-h)^2 + (y-k)^2 = r^2\). In this form:

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

By rewriting an equation into this standard form, you can easily identify these characteristics. This is because the expression \((x-h)^2 + (y-k)^2\) describes the squared distance from any point \((x, y)\) on the circle to the center \((h, k)\). Thus, an equation that initially does not seem to represent a circle can be transformed through completing the square.
Once in the standard form, it clearly shows the geometrical nature of the circle, making it a powerful tool in solving problems related to circles.

Center and Radius of a Circle

The center and the radius are defining properties of a circle. In an equation of the form \((x-h)^2 + (y-k)^2 = r^2\), it becomes straightforward to extract the circle's fundamental details.

• The center

is represented by the coordinate \((h, k)\). This is where the circle is positioned in a plane.

• The radius

is given by \(r\) in the equation. It's determined by taking the square root of the number on the right side of the equation, \(r^2\). Knowing how to identify these two values from the equation is essential.
Whether you're plotting the circle on a graph or analyzing its properties, the center and radius provide vital information about its size and position within coordinate geometry.