Problem 14 - The Calculus with Analytic Geometry

Question

In Exercises 11 through 14 , find the center and radius of each circle, and draw a sketch of the graph. $$ x^{2}+y^{2}-10 x-10 y+25=0 $$

Step-by-Step Solution

Answer

The center is (5, 5) and the radius is 5.

Step 1 - Write the General Form of a Circle Equation

The general equation of a circle is \[ (x-h)^2 + (y-k)^2 = r^2 \]. We need to transform the given equation into this form.

Step 2 - Group and Complete the Square

Group the x and y terms together:\[ x^2 - 10x + y^2 - 10y + 25 = 0 \]. Complete the square for the x and y terms:\[ x^2 - 10x = (x-5)^2 - 25 \]and\[ y^2 - 10y = (y-5)^2 - 25 \].

Step 3 - Substitute Back into the Equation

Rewrite the original equation with the completed squares:\[ (x-5)^2 - 25 + (y-5)^2 - 25 + 25 = 0 \]. Combine like terms:\[ (x-5)^2 + (y-5)^2 - 25 = 0 \]. Then rearrange to:\[ (x-5)^2 + (y-5)^2 = 25 \].

Step 4 - Identify the Center and Radius

From the equation \[ (x-5)^2 + (y-5)^2 = r^2 \] we see that \[ h = 5 \] and \[ k = 5 \] so the center is \[ (5, 5) \]. Also, \[ r^2 = 25 \] so the radius is \[ r = 5 \].

Step 5 - Draw the Circle

Draw a circle with center at (5, 5) and a radius of 5 units. This sketch will help visualize the solution.

Key Concepts

Completing the Square

Completing the square is a method used to transform a quadratic expression into a perfect square trinomial. This technique is particularly useful for converting the general form of a circle's equation into the standard form. First, we start with the general equation of the circle:

\[ x^2 + y^2 - 10x - 10y + 25 = 0 \]

We need to rearrange the equation to make completing the square easier. Group the x terms and y terms together, and then isolate the constant term:

\[ x^2 - 10x + y^2 - 10y = -25 \]

Now, let's complete the square for the x and y terms. For the x terms:

1. Take the coefficient of x, which is -10, divide it by 2, and square the result:
\[ \bigg(\frac{-10}{2}\bigg)^2 = 25 \]

2. Add and subtract 25 within the x terms:
\[ x^2 - 10x = (x-5)^2 - 25 \]

Repeat the same process for the y terms:

1. Take the coefficient of y, which is -10, divide it by 2, and square the result:
\[ \bigg(\frac{-10}{2}\bigg)^2 = 25 \]

2. Add and subtract 25 within the y terms:
\[ y^2 - 10y = (y-5)^2 - 25 \]

Finally, substitute these completed squares back into the equation.

General Form of a Circle Equation

The general equation of a circle is given by \[ (x-h)^2 + (y-k)^2 = r^2 \]

Here, - \( (h, k) \) represents the center of the circle - \( r \) is the radius of the circle

Let's return to our equation after Completing the Square: \[ (x-5)^2 - 25 + (y-5)^2 - 25 + 25 = 0 \]

Simplify by combining like terms:

\[ (x-5)^2 + (y-5)^2 - 25 + 25 = 0 \]
\[ (x-5)^2 + (y-5)^2 = 25 \]

Now, the equation is in the form \[ (x-h)^2 + (y-k)^2 = r^2 \].

### Identifying the Center and Radius

From this, it's clear that
- h = 5, k = 5 which gives the center (5, 5)
- \( r^2 = 25 \) so the radius r = 5.

This simplified form helps us easily identify the circle's key components.

Geometry of a Circle

In geometry, a circle is defined as the set of all points in a plane that are at a fixed distance from a given point called the center. The fixed distance is known as the radius.

### Sketching the Circle

Once you have the center and the radius, you can easily sketch the graph of the circle. For the given equation \[ (x-5)^2 + (y-5)^2 = 25 \],

we have:

- Center at (5, 5)- Radius of 5 units

To draw the circle:

1. Plot the center at (5, 5).2. Using a compass, set the distance to 5 units.3. Draw a circle around the center spanning 5 units in all directions.

This visual representation helps solidify understanding of the circle's properties. By recognizing and using both the algebraic and geometric interpretations of a circle, students can gain a deeper comprehension of the topic.