Problem 5
Question
A. Write the standard form of the equation of a circle. B. Write the standard form of the equation of a circle with the center at the origin.
Step-by-Step Solution
Verified Answer
A: \((x - h)^2 + (y - k)^2 = r^2\); B: \(x^2 + y^2 = r^2\).
1Step 1: Understanding Standard Form of a Circle Equation
The standard form of a circle's equation is derived from the distance formula. It is expressed as \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center of the circle, and \(r\) is the radius.
2Step 2: Center of Circle at Origin
When the center of the circle is at the origin, the coordinates \((h, k)\) are \((0, 0)\). This simplifies the equation to \((x - 0)^2 + (y - 0)^2 = r^2\).
3Step 3: Simplifying the Equation with Center at Origin
With \((h, k) = (0, 0)\), the equation becomes \(x^2 + y^2 = r^2\). This is the standard circular equation centered at the origin.
Key Concepts
Distance FormulaCircle EquationCenter of a Circle
Distance Formula
The distance formula is a vital tool in understanding the geometry of circles, among other shapes. It helps to determine the distance between two points in a plane. The formula is given by:
\[\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]This formula is derived from the Pythagorean theorem and calculates the straight line distance between the points \((x_1, y_1)\) and \((x_2, y_2)\).
\[\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]This formula is derived from the Pythagorean theorem and calculates the straight line distance between the points \((x_1, y_1)\) and \((x_2, y_2)\).
- Used to find the radius of a circle, by measuring from the center to a point on the circle.
- Essential in developing the standard form equation of a circle by substituting the center and a point on the circle.
Circle Equation
The circle equation in its standard form is an elegant representation of all points equidistant from a central point in a plane. The standard form is:
\[(x - h)^2 + (y - k)^2 = r^2\]Here, \((h, k)\) defines the center of the circle and \(r\) is the radius. This equation is powerful in geometry because it encapsulates all necessary information about the circle in a simple format.
\[(x - h)^2 + (y - k)^2 = r^2\]Here, \((h, k)\) defines the center of the circle and \(r\) is the radius. This equation is powerful in geometry because it encapsulates all necessary information about the circle in a simple format.
- The terms \((x - h)^2\) and \((y - k)^2\) ensure every point on the circle is the same distance, \(r\), from the center.
- The square format is derived from the distance formula, emphasizing the constant radius.
Center of a Circle
The center of the circle is the referential point from which all points on the edge of the circle are equidistant. Given the standard form equation \((x - h)^2 + (y - k)^2 = r^2\), the center is represented by \((h, k)\).
Understanding the center is crucial because:
Understanding the center is crucial because:
- If the center is \((0, 0)\), it greatly simplifies the equation to \(x^2 + y^2 = r^2\), creating a neat symmetry around the origin.
- The role of \(h\) and \(k\) is to shift the circle across the plane without altering its shape, merely changing its position.
Other exercises in this chapter
Problem 5
Fill in the blanks. A _____ is a line that intersects a circle at two points.
View solution Problem 5
Fill in the blanks. The line segment joining the vertices of an ellipse is called the ________ axis of the ellipse.
View solution Problem 6
Fill in the blanks. A _____ is a line that intersects a circle at one point.
View solution Problem 6
Fill in the blanks. To write \(9 x^{2}-4 y^{2}=36\) in ______ form, we divide both sides by 36
View solution