Problem 63 - Algebra 2

Question

Find the center and the radius of each circle. $$ x^{2}+y^{2}-6 x-2 y+4=0 $$

Step-by-Step Solution

Answer

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

Step 1: Rearrange the equation into the standard form

First, group the \(x\) terms and \(y\) terms together, giving us the following equation: \(x^{2}-6x + y^{2}-2y = -4\).

Step 2: Complete the square for \(x\) and \(y\)

Complete the square for the \(x\) terms by adding \((-6/2)^{2}\) to both sides and the \(y\) terms by adding \((-2/2)^{2}\) to both sides. This gives us \(x^{2}-6x +9 + y^{2}-2y +1 = -4 + 9 + 1\).

Step 3: Simplify the equation

The equation can be simplified to \((x-3)^{2}+(y-1)^{2}=6\).

Step 4: Identify the center and radius

In the standard form of the circle equation, the center is at (\(h\), \(k\)) and the radius is \(\sqrt{r^{2}}\). So, for our simplified equation, the center is at (3, 1) and the radius is \(\sqrt{6}\).

Key Concepts

Completing the Square

Completing the square is a method used to simplify quadratic equations. The goal is to form a perfect square trinomial, which is a three-term expression that can be factored into a square of a binomial.
For example, let's take a look at the terms involving \(x\) in the given equation \(x^{2} - 6x\). To complete the square:

• Take the coefficient of \(x\), which is -6, and divide it by 2, yielding -3.
• Then square it, resulting in 9.
• Add this 9 to both sides of the equation to maintain equality.

This transforms \(x^{2} - 6x\) into \((x - 3)^{2}\).
The same process is applied to the \(y\) terms. For \(y^{2} - 2y\):

• The coefficient is -2. Halve it and square it, giving 1.
• Add 1 to both sides.

This changes \(y^{2} - 2y\) into \((y - 1)^{2}\).
By completing the square, the equation becomes easier to interpret and leads us to the geometrical representation of a circle.

Center of a Circle

The center of a circle in its standard equation form, \((x-h)^{2} + (y-k)^{2} = r^{2}\), is given as the point \((h, k)\). This represents the circle's central point from which all points on the circumference are equidistant.
When we transformed our original circle equation by completing the square, it became \((x - 3)^{2} + (y - 1)^{2} = 6\). From this:

• The coefficient of the transformed \(x\) term, which was obtained from \(x^{2} - 6x\), is \(h = 3\).
• Similarly, for the transformed \(y\) term, derived from \(y^{2} - 2y\), is \(k = 1\).

Therefore, the center of the circle is the point \((3, 1)\).
This point is key when drawing the circle or determining its position relative to other shapes or coordinate points. Understanding the center helps in exploring various mathematical properties and relationships.

Radius of a Circle

The radius of a circle is the distance from its center to any point on its boundary. In a circle's standard equation \((x - h)^{2} + (y - k)^{2} = r^{2}\), the value of \(r\) represents the radius squared.
Once we completed the square, the equation changed to \((x - 3)^{2} + (y - 1)^{2} = 6\). Here:

• The value 6 on the right-hand side represents \(r^{2}\). To find the actual radius (\(r\)), we need to take the square root of 6.
• Thus, the radius is \(\sqrt{6}\).

Understanding how to derive the radius from the equation is important for determining the size of a circle. It's used both in practical applications and theoretical math.
Knowing the radius, alongside the center, allows you to recreate the shape and examine its properties effectively.