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Chapter 13: Problem 15

Find the center and radius of each circle. (Hint: In Exercises 15 and \(16,\) divide each side by a common factor) $$ 3 x^{2}+3 y^{2}-12 x-24 y+12=0 $$

Short Answer

Expert verified
The center is (2, 4) and the radius is 4.

Step by step solution

01

Simplify the equation

Divide each term in the equation by 3 to make the equation simpler:\[ \frac{3x^2 + 3y^2 - 12x - 24y + 12}{3} = 0 \]This simplifies to:\[ x^2 + y^2 - 4x - 8y + 4 = 0 \]
02

Rearrange the equation

Move the constant term to the other side of the equation:\[ x^2 + y^2 - 4x - 8y = -4 \]
03

Complete the square for x

To complete the square for the x terms, take half of the coefficient of x (-4), square it, and add it inside the equation:\[ x^2 - 4x \rightarrow (x - 2)^2 - 4 \]So the x terms become:\[ (x - 2)^2 - 4 \]
04

Complete the square for y

Similarly, take half of the coefficient of y (-8), square it, and add it inside the equation:\[ y^2 - 8y \rightarrow (y - 4)^2 - 16 \]So the y terms become:\[ (y - 4)^2 - 16 \]
05

Combine completed squares

Include the changes from steps 3 and 4 in the equation. It now looks as follows:\[ (x - 2)^2 - 4 + (y - 4)^2 - 16 = -4 \]Combine constants on the right side of the equation:\[ (x - 2)^2 + (y - 4)^2 - 20 = -4 \]When simplified, this turns into:\[ (x - 2)^2 + (y - 4)^2 = 16 \]
06

Identify the center and the radius

The equation \[ (x - 2)^2 + (y - 4)^2 = 16 \] represents a circle with center at (2, 4) and radius \( \text{r} = \text{sqrt}(16) = 4 \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

completing the square
To transform the equation of a circle into its standard form, we use a method called completing the square. This method helps us recognize the circle's radius and center by restructuring the equation into a more familiar format. Suppose we have an expression like \(x^2 - 4x\). Here鈥檚 how you complete the square for it:

1. Take the coefficient of \(x\) (which is -4), divide it by 2, and square the result.
2. \((-4 / 2) ^ 2 = 4\)
3. Add and subtract this square value within the equation to balance it.

This allows us to rewrite the quadratic part of the equation as a perfect square trinomial, making it easier to handle and solve for circle properties.
circle center
The center of a circle in its standard form equation \((x - h)^2 + (y - k)^2 = r^2\) is represented by the point (h, k). In our example equation \((x - 2)^2 + (y - 4)^2 = 16\), after completing the square, the center is identified as (2, 4).
The process involves:

- Completing the square for both x and y terms.
- Identifying the constants \((x - h)\) and \((y - k)\).
Once in standard form, simply read off the values of h and k to determine the center of the circle.
circle radius
To find the radius of the circle, we look at the standard equation \((x - h)^2 + (y - k)^2 = r^2\). The value r^2 on the right side provides the radius squared. To get the radius, we take the square root of this value. In our example, the equation after transformation is \((x - 2)^2 + (y - 4)^2 = 16\).
Simplify 16 as follows:
  • Square root of 16 is 4.

      • Therefore, the radius of our circle is 4 units.
standard form of a circle
The standard form of a circle's equation is written as \((x - h)^2 + (y - k)^2 = r^2\). This shows all necessary information about the circle: its center (h, k) and its radius \(r\).

Transforming an equation like \(3x^2 + 3y^2 - 12x - 24y + 12 = 0\) involves factoring out constants, rearranging, and completing the square:
  • Factor out constants to simplify.

  • Move constants to the other side of the equation.

    Once in the standard form, it is straightforward to extract the circle's properties, providing a neat and clean equation illustrating both center and radius.

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Most popular questions from this chapter

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