Problem 38
Question
Graph each circle so that the center is at the origin. Then write the equation. diameter 2.5
Step-by-Step Solution
Verified Answer
The equation of the circle is \( x^2 + y^2 = 1.5625 \). Note: The plotted circle could vary depending on the type of graphing tools used.
1Step 1: Determine radius
The radius of a circle is half of its diameter. Given the diameter is 2.5, the radius of the circle is \( \frac{2.5}{2} = 1.25 \).
2Step 2: Write circle equation
Use the general equation of a circle \( x^2 + y^2 = r^2 \) . Substituting the radius value calculated above, the equation of the circle becomes \( x^2 + y^2 = (1.25)^2 = 1.5625 \).
3Step 3: Plot the circle
On a graph, plot the circle with its center at the origin (0,0). The circle should pass through the points at which x and y coordinates are 1.25 or -1.25, due to its radius being 1.25. Hence, four points where circle can pass through are (1.25,0), (-1.25,0), (0,1.25) and (0,-1.25).
Key Concepts
Radius and DiameterPlotting GraphsCenter at the Origin
Radius and Diameter
The radius and diameter are fundamental concepts when working with circles. The diameter is the distance across the circle, passing through the center. The radius, on the other hand, is half the diameter. It extends from the center to any point on the circle's boundary. In mathematical terms:
- Diameter: The full width of the circle, which is always twice the radius.
- Radius: Half the diameter, calculated by dividing the diameter by two.
Plotting Graphs
Plotting graphs is a way to visualize mathematical concepts, like circles. Start by finding key points that lie on the circle. This begins with understanding that the center is at the origin, which is the point (0,0). From there, use the radius to identify boundary points. For a circle with a radius of 1.25, it will touch the graph at several distinct locations:
- (1.25, 0)
- (-1.25, 0)
- (0, 1.25)
- (0, -1.25)
Center at the Origin
When a circle's center is at the origin, its mathematical equation becomes simplified. The origin is the point (0,0) on a graph where the x-axis and the y-axis intersect. For circles, this point is often used as the center because it simplifies calculations and helps in understanding symmetrical properties. The general equation for a circle with its center at the origin is given by:\[ x^2 + y^2 = r^2 \]Here, \(r\) is the radius. So, for a radius of 1.25, we plug this value into the equation as:\[ x^2 + y^2 = (1.25)^2 = 1.5625 \]This equation shows that every point \((x, y)\) on the circle is equidistant from the origin, with that distance being the circle's radius. Setting the center at the origin also assists with plotting since it aligns the circle symmetrically across the axes.
Other exercises in this chapter
Problem 38
The graph of each equation is to be translated 3 units right and 5 units up. Write each new equation. \(\frac{(x-3)^{2}}{64}+\frac{(y+3)^{2}}{36}=1\)
View solution Problem 38
List the properties of a hyperbola that allow you to sketch its graph.
View solution Problem 38
Write an equation of a parabola with a vertex at the origin. \(\operatorname{directrix} y=5\)
View solution Problem 38
Write the equation of the circle that passes through the given point and has a center at the origin. (Hint: You can use the distance formula to find the radius.
View solution