Problem 29 - Beginning and Intermediate Algebra

Question

Put the equation of each circle in the form \((x-h)^{2}+(y-k)^{2}=r^{2},\) identify the center and the radius, and graph. $$x^{2}+y^{2}+2 x+10 y+17=0$$

Step-by-Step Solution

Answer

The given equation \(x^{2}+y^{2}+2 x+10 y+17=0\) doesn't represent a circle in the standard form, after completing the square, we obtain the equation \((x+1)^{2}+(y+5)^{2}=-41\). Since we can't have a negative radius, there is no center, radius, or graph corresponding to this equation.

Step 1: Rewrite the equation in standard form

To rewrite the given equation in the standard form, use the completing the square method. Given equation: \(x^{2}+y^{2}+2 x+10 y+17=0\) Group the x and y terms together: \((x^{2}+2 x)+(y^{2}+10 y)=-17\) Now, complete the square for x and y terms: For x: Take half of the x's coefficient (half of 2 is 1), square it (1^2 = 1), and add it inside the parenthesis. But don't forget to subtract the added value on the other side of equation. For y: Take half of the y's coefficient (half of 10 is 5), square it (5^2 = 25), and add it inside the parenthesis. But don't forget to subtract the added value on the other side of equation. The equations become: \((x^{2}+2 x+1)+(y^{2}+10 y+ 25)=-17+1-25\) Now, rewrite the equation as: \((x+1)^{2}+(y+5)^{2}=-41\) But, since we can't have a negative radius, the given equation can't represent any circle.

Step 2: Center and Radius

Since we have concluded that the given equation doesn't represent a circle, we can't find the center and radius.

Step 3: Graphing the Circle

As this equation doesn't represent a valid circle, we can't graph it. The given equation doesn't represent a circle in the standard form, so there is no center, radius, or graph corresponding to this equation.

Key Concepts

Standard Form of a Circle

The standard form of a circle is expressed through the equation \((x-h)^{2}+(y-k)^{2}=r^{2}\). This represents a circle in a coordinate plane.
Here, the

• "\((x-h)\)" and "\((y-k)\)" are used to translate the center of the circle to any point \((h,k)\) on the plane.
• "\(r^{2}\)" is the square of the radius of the circle.

To put an equation into this form, we often need to employ the method of completing the square, particularly when dealing with quadratic terms.
This technique allows us to rearrange the given equation into a form that clearly shows the circle's center and radius.

Equation of a Circle

When we talk about the equation of a circle, we're referring to the connection between all points on the circle and its center. The standard equation \((x-h)^{2}+(y-k)^{2}=r^{2}\) helps us to precisely describe this relationship. If there is interference from additional terms, like in the case of our exercise with terms outside \(x, y\) or constant terms, completing the square comes to the rescue.

• Completing the square involves taking a trinomial and transforming it into a perfect square trinomial.
• For example, from \(x^2 + 2x\), we take half of 2, square it, and add 1 to complete the square.
• This step transforms our trinomial into \((x+1)^2\).

This method shows a pivotal role in rearranging initial circle equations because it reveals the actual center and radius.
The arrangement determines if the equation is valid—negative results, like in our original problem, mean the equation cannot form a real circle.

Graphing Circles

Graphing a circle provides a visual representation of the circle's equation. Once we have an equation in standard form \((x-h)^{2}+(y-k)^{2}=r^{2}\), it's simple to plot:

• Begin by identifying the center \((h,k)\).
• Then, from this center, measure the radius \(r\) in all directions (up, down, left, right) to plot the circle.

However, if the equation yields a negative value, like \(-41\) in the original problem, it tells us that there is no circle to graph.
This teaches us about the intricacies of equations and mathematical representation—if the radius squared \(r^{2}\) becomes negative, we understand that something does not exist in the visible, two-dimensional realm.
Graphing helps confirm our calculations and visually verify the center and radius, reinforcing the connection between algebraic manipulation and visual geometry.