Problem 55 - Precalculus Essentials

Question

Complete the square and write the equation in standard form. Then give the center and radius of each circle and graph the equation. $$x^{2}+y^{2}-10 x-6 y-30=0$$

Step-by-Step Solution

Answer

The standard form of the given equation is \((x-5)^{2} + (y-3)^{2} = 64\). The center of the circle is at point (5,3) and the radius is 8 units.

Step 1: Rewrite equation in standard form

Begin by grouping the x terms and y terms separately: \(x^{2}-10x+y^{2}-6y = 30\). Add the square of half of coefficient of x and y to both sides to complete the squares: \((x^{2}-10x+25) + (y^{2}-6y+9) = 30 + 25 + 9\). This gives us the equation in standard form \((x-5)^{2} + (y-3)^{2} = 64\).

Step 2: Identify the center and radius

From the standard form, the center of the circle can be read-off as (5,3). The radius is the square root of 64, therefore, the radius, r=8.

Step 3: Graph the equation

To graph the equation, first plot the center (h,k) = (5,3) on the x-y plane. Next, draw a circle with a radius of 8 units. The resulting graph represents the given equation.

Key Concepts

Standard Form of a Circle

Understanding the standard form of a circle is crucial for solving problems related to circular shapes in coordinate geometry. The equation of a circle in standard form is given by \( (x-h)^{2} + (y-k)^{2} = r^{2} \), where \( (h,k) \) represents the center of the circle and \( r \) is the radius.

Let's break it down using the example from the exercise. The given equation \( x^{2}+y^{2}-10x-6y-30=0 \) does not immediately reveal the properties of the circle. However, by completing the square, we transform it into a recognizable form that conveys meaningful information about the circle's dimensions and position.

Completing the square involves grouping the \( x \) and \( y \) terms and then adding and subtracting the same value to preserve the equation's equality. This manipulation constructs perfect square trinomials from the \( x \) and \( y \) terms, hence 'completing the square.' The resulting standard form, \( (x-5)^{2} + (y-3)^{2} = 64 \), not only makes it easy to identify the circle's characteristics but also serves as a building block for further analyses and applications, such as finding intersections with other geometrical shapes or optimizing areas.

Center and Radius of a Circle

Now let's delve into identifying the center and radius of a circle, which directly stem from its standard form equation. The equation \( (x-5)^{2} + (y-3)^{2} = 64 \) from the example provided gives us this vital information in a transparent manner.

The coordinates \( h \) and \( k \) inside the squared terms represent the center of the circle. In this case, the center is \( (5, 3) \). To visualize why, imagine moving along the x-axis to 5 and then up the y-axis to 3; that point marks the center around which the circle expands uniformly.

The radius can be found by taking the square root of the constant on the right side of the equation, which in our example is 64. Thus, the radius \( r \) is 8 units. The radius gives us a measure of how big the circle is and it tells us how far from the center any point on the circle lies. Remember, all points on the edge of the circle are exactly \( r \) units from the center, creating the boundary of this perfectly symmetrical shape.

Graphing Equations

Lastly, we'll cover graphing equations, specifically our circle. With the equation in standard form and knowledge of the center and radius, graphing becomes a straightforward task.

To plot our circle \( (x-5)^{2} + (y-3)^{2} = 64 \) on the coordinate plane, we start by marking the center at the point \( (5, 3) \). From the center, we draw a circle with a constant distance, the radius, from this point. That distance is 8 units in all directions.

While drawing the circle freehand, it's helpful to mark points on the coordinate plane that are 8 units away from the center along the cardinal directions—up, down, left, and right. Then, using those points as a guide, sketch the circle. If possible, use a compass for accuracy. Graphing the circle accurately presents the visual representation of the equation and can serve as a powerful tool for understanding relationships between different geometrical entities within the coordinate system.