Problem 38 - 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. $$x^{2}+y^{2}-10 y+22=0$$

Step-by-Step Solution

Answer

The standard form of the circle equation is \((x-0)^2+(y-5)^2=47\), with the center at (0,5) and the radius equal to \(\sqrt{47}\). To graph the circle, plot the center at (0,5) and draw a circle with radius \(\sqrt{47}\).

Step 1: Write down the given equation

We have the given equation: \(x^2+y^2-10y+22=0\)

Step 2: Complete the square for the y-term

To complete the square for the y-term, we need to find the value that needs to be added on both sides of the equation. Divide the coefficient of the y-term by 2 and then square it. \(\left(\frac{-10}{2}\right)^2 = (-5)^2 = 25\) Now, add this value to both sides of the equation: \(x^2 + y^2-10y + 25 = 22+25\)

Step 3: Simplify and put the equation in standard form

By simplifying the equation, we can put it in the standard form \((x-h)^{2}+(y-k)^{2}=r^{2}\): \(x^2 +(y-5)^2 = 47\) Now, we have the equation in standard form.

Step 4: Identify the center and radius

Using the standard form, \((x-h)^2 + (y-k)^2 = r^2\), we can identify the center and radius of the circle. The center of the circle is (h,k) = (0,5) and the radius is \(r = \sqrt{47}\).

Step 5: Graph the circle

Plot the center of the circle (0,5) on the coordinate plane. Using the radius, draw a circle with the center (0,5) and radius equal to \(\sqrt{47}\). Ensure that the circle is complete and represents the equation properly. Combining all these steps, we now have the standard form of the circle equation \((x-0)^2+(y-5)^2=47\), with center (0,5) and radius \(\sqrt{47}\). The graph should show a circle, centered at (0,5) and with a radius equal to \(\sqrt{47}\).

Key Concepts

Completing the Square

Completing the square is a technique used to transform a quadratic expression into a perfect square trinomial. This helps in rewriting quadratic equations into a form that reveals key characteristics, such as the center and radius of a circle.
To complete the square, identify the quadratic term that needs conversion - often, this is applied to terms involving variables such as \(y\) in our exercise.
First, take the coefficient of the middle term (e.g., \( -10y \)) and divide it by 2. Then, square the result.

• For \(-10y\), dividing by 2 gives \(-5\).
• Squaring \(-5\) gives 25.

Add 25 to both sides of the equation to maintain equality. This transforms the terms into a perfect square trinomial, thus allowing you to factor it easily into \((y-5)^2\).
Completing the square reorganizes your equation into a pattern that closely resembles part of a circle's standard form.

Standard Form of a Circle

The standard form of a circle's equation is a valuable form as it readily shows the circle's key characteristics. The standard form is written as:\[(x-h)^2 + (y-k)^2 = r^2\]This version of the equation allows you to easily locate the center of the circle and calculate its radius.
By completing the square for any necessary terms and rearranging the original equation, you can transform it into this standard form.
As shown in the exercise, the original equation \(x^2+y^2-10y+22=0\), after completing the square for \(y\), becomes \((x-0)^2 + (y-5)^2 = 47\).

• Key benefit: The form clearly presents the parameters needed to identify the circle's key features without additional manipulation or calculation.

Identifying Center and Radius

Identifying the center \((h, k)\) and radius \(r\) of a circle from its equation in standard form is direct once you have the equation organized.
The \(h\) and \(k\) represent shifts from the origin in the horizontal and vertical directions, respectively.

• In the form \((x-h)^2 + (y-k)^2 = r^2\), the center of the circle is \((h, k)\).
• For our equation, these values are \(0\) and \(5\), giving us a center at \((0, 5)\).

The radius \(r\) is then calculated by taking the square root of the number on the right of the equation. It's important to remember that the radius should be a positive value.

• Here, \(r^2 = 47\), so the radius \(r = \sqrt{47}\).

With the center and the radius in hand, one can accurately visualize and plot the circle on a coordinate plane.