Problem 57
Question
complete the square and write the equation in standard form. Then give the center and radius of each circle and graph the equation. $$ x^{2}+y^{2}+8 x-2 y-8=0 $$
Step-by-Step Solution
Verified Answer
The center of the circle is (-4, 1) and its radius is 5.
1Step 1: Rearrange the equation grouping x's and y's together
Rearrange the equation of the circle such that all terms involving x are together, all terms involving y are together, and the constant term is on the other side, which gives \( x^{2} + 8 x + y^{2} - 2y = 8 \)
2Step 2: Complete the square for x and y
For completing the square, you need to add and subtract the square of half the coefficient of x and y inside their respective brackets. This yields: \((x^{2} + 8x + 16) -(16) + (y^{2} - 2y + 1) - (1) = 8 \) Which simplifies to: \((x + 4)^{2} - 16 + (y - 1)^{2} - 1 = 8 \)
3Step 3: Simplify the equation to a standard form
Simplify the equation to put it in a standard form. Here, we combine all the constant terms on the right-hand side of the equation, so this will give: \((x + 4)^{2} + (y - 1)^{2} = 25 \)
4Step 4: Identify the center and radius
The equation of the circle in standard form is \( (x - a)^{2} + (y - b)^{2} = r^{2} \). Comparing this with our equation \((x + 4)^{2} + (y - 1)^{2} = 5^{2}\), you can see that a = -4, b = 1, and r = 5. Therefore, the center of the circle is (-4, 1) and the radius is 5.
5Step 5: Graph the circle
Plot the center of the circle (-4,1) on the graph. Then, move 5 units (radius) away from the center in all directions and make the circle.
Key Concepts
Standard Form of Circle EquationCenter of a CircleRadius of a CircleGraphing Circles
Standard Form of Circle Equation
The standard form of a circle equation is a neat way to express circular shapes in coordinate geometry. It is crucial for identifying the center and radius of a circle easily. This form is written as:\[(x - a)^2 + (y - b)^2 = r^2\]Here,
- \((x - a)^2\) and \((y - b)^2\) are terms derived through a method called completing the square.
- \(a\) and \(b\) represent the coordinates of the circle's center.
- \(r\) is the radius, which tells us how far the circle stretches from its center.
Center of a Circle
The center of a circle is a pivotal point, often symbolized by coordinates in a plane. In the formula \((x - a)^{2} + (y - b)^{2} = r^{2}\), the values
In our example, after rewriting the equation to standard form,
- \(a\), and
- \(b\)
In our example, after rewriting the equation to standard form,
- the center lands at the point \((-4, 1)\).
Radius of a Circle
A circle's radius is the constant distance from its center to any point along its edge. In the equation \[(x - a)^2 + (y - b)^2 = r^2\]\(r\) denotes the radius. It is determined by simplifying the terms linked with the standard form. In this instance, through completing the square and rearranging, we found
- \(r = 5\)
Graphing Circles
Graphing a circle involves translating the equation of a circle on a coordinate plane visually. After determining the standard form, center, and radius:
- Begin by plotting the center of the circle, which is \((-4, 1)\) in our example.
- Then, use the radius, here it is 5, to draw the curve. Move \(5\) units in each direction (up, down, left, and right) from the center to ensure accurate placement.
- Finally, sketch the circle by connecting these boundary points smoothly in a round fashion.
Other exercises in this chapter
Problem 56
Begin by graphing the standard quadratic function, \(f(x)=x^{2} .\) Then use transformations of this graph to graph the given function. $$g(x)=(x-1)^{2}$$
View solution Problem 56
Graph each equation in a rectangular coordinate system. $$f(x)=3$$
View solution Problem 57
find and simplify the difference quotient $$ \frac{f(x+h)-f(x)}{h}, h \neq 0 $$ for the given function. $$ f(x)=3 x+7 $$
View solution Problem 57
Find a. \((f \circ g)(x) \quad \) b. \((g \circ f)(x) \quad \) c. \((f \circ g)(2) \quad \) d. \((g \circ f)(2)\) $$f(x)=x^{2}+2, g(x)=x^{2}-2$$
View solution