Problem 84 - Intermediate Algebra

Question

Find the center and radius of each circle. A. \(x^{2}+y^{2}+10 x-14 y-7=0\) B. \(x^{2}+y^{2}-10 x+14 y-7=0\)

Step-by-Step Solution

Answer

A: Center (-5, 7), Radius 9; B: Center (5, -7), Radius 9.

Step 1: Rearrange the Circle Equation

First, take the given equation of the circle: A. \(x^2 + y^2 + 10x - 14y - 7 = 0\) Our goal is to rewrite this in standard form: \((x-h)^2 + (y-k)^2 = r^2\). We start by grouping the terms: \((x^2 + 10x) + (y^2 - 14y) = 7\)

Step 2: Complete the Square for x

For the x-terms \(x^2 + 10x\), we complete the square. Take half of the coefficient of \(x\) (which is 10), square it, and add that number inside the brackets:\(10 / 2 = 5; \quad 5^2 = 25\)Add and subtract 25 inside the equation:\((x^2 + 10x + 25 - 25) + (y^2 - 14y) = 7\)Thus, \((x+5)^2 - 25\) becomes part of the expression.

Step 3: Complete the Square for y

For the y-terms \(y^2 - 14y\), we complete the square in a similar manner. Take half of the coefficient of \(y\) (which is -14), square it, and add that number inside the brackets:\(-14 / 2 = -7; \quad (-7)^2 = 49\)Add and subtract 49 inside the equation:\((x+5)^2 - 25 + (y^2 - 14y + 49 - 49) = 7\)This simplifies to:\((x+5)^2 + (y-7)^2 - 25 - 49 = 7\)\((x+5)^2 + (y-7)^2 = 81\)

Step 4: Determine Center and Radius

The equation \((x+5)^2 + (y-7)^2 = 81\) is now in standard form. The center \((h, k)\) of the circle is \((-5, 7)\), and the radius \(r\) is the square root of 81, which is \(9\).

Step 5: Repeat for the Second Equation

Now, repeat Steps 1 through 4 for the second equation:B. \(x^2 + y^2 - 10x + 14y - 7 = 0\)Rearrange into the format \((x^2 - 10x) + (y^2 + 14y) = 7\).

Step 6: Complete the Square for x in Second Equation

Complete the square for \(x\):\(-10 / 2 = -5; \quad (-5)^2 = 25\)\((x^2 - 10x + 25 - 25) = (x-5)^2 - 25\)

Step 7: Complete the Square for y in Second Equation

Complete the square for \(y\):\(14 / 2 = 7; \quad 7^2 = 49\)\((y^2 + 14y + 49 - 49) = (y+7)^2 - 49\)

Step 8: Final Reformat and Calculate Center/Radius for Second Equation

Combine results:\((x-5)^2 + (y+7)^2 - 25 - 49 = 7\)\((x-5)^2 + (y+7)^2 = 81\)The center for this circle is \((5, -7)\), and the radius is also \(9\) as in part A.

Key Concepts

Completing the Square

Completing the square is a technique used in algebra to transform a quadratic expression into a perfect square trinomial. This method is very helpful when rewriting circle equations into the standard form. Here's how it works:
When you have a quadratic term like \(x^2 + bx\), you can complete the square by following these steps:

• First, take the coefficient of \(x\), which we call \(b\).
• Divide \(b\) by 2. This gives you \(\frac{b}{2}\).
• Square the result of \(\frac{b}{2}\). This means you calculate \(\left(\frac{b}{2}\right)^2\).
• Add and subtract this square inside the expression. This allows you to write the trinomial as a perfect square.

For example, with the expression \(x^2 + 10x\):
- Divide 10 by 2 to get 5.
- Square 5 to get 25.
You then express \(x^2 + 10x\) as \((x+5)^2 - 25\).
This completes the square for the \(x\)-terms, turning the expression into a neat square-format. The same approach applies to the \(y\)-terms when needed.

Standard Form of a Circle

The standard form of a circle's equation provides a clear way to identify its center and radius, which are fundamental properties of a circle. The standard form equation looks like this:

• \((x-h)^2 + (y-k)^2 = r^2\)

In this equation:

• \(h\) and \(k\) represent the x and y coordinates of the center of the circle.
• \(r\) is the radius of the circle.

For example, by rearranging and completing the square for the given equation \(x^2 + y^2 + 10x - 14y - 7 = 0\), we eventually rewrite it in the standard form:
\((x+5)^2 + (y-7)^2 = 81\).
This layout immediately tells us both the center and the radius of the circle.

Circle Center and Radius

Discovering the center and radius from the equation of a circle is simplified once the equation is in standard form. From \((x-h)^2 + (y-k)^2 = r^2\):

• The point \((h, k)\) is the center of the circle.
• The radius, \(r\), is found by taking the square root of the right side.

In the exercise given, after completing the square, we obtained:

• For Equation A: \((x+5)^2 + (y-7)^2 = 81\)
• The center is located at \((-5, 7)\).
  
• The radius is \(\sqrt{81} = 9\).

In the second equation, after completing the same process, we have:

• For Equation B: \((x-5)^2 + (y+7)^2 = 81\)
• The center is \((5, -7)\).
• The radius remains \(9\).

This universal approach makes identifying key features of a circle straightforward and clear once standard form is achieved.