Problem 65

Question

\(65-72\) . Show that the equation represents a circle, and find the center and radius of the circle. $$ x^{2}+y^{2}-2 x+4 y+1=0 $$

Step-by-Step Solution

Verified
Answer
Center is (1, -2); Radius is 2.
1Step 1: Rewrite the Circle Equation
Start by rewriting the equation in a form that resembles a circle equation: \ \[ x^2 + y^2 - 2x + 4y + 1 = 0 \] \ Move the constant term to the other side: \ \[ x^2 - 2x + y^2 + 4y = -1 \]
2Step 2: Complete the Square for x-terms
To complete the square for the \(x\) terms, take the coefficient of \(x\), divide by 2, and square it. \ The coefficient of \(x\) is -2. Dividing by 2 gives -1, and squaring gives 1. \ Rewrite the \(x\) terms: \ \[ (x^2 - 2x) \rightarrow (x - 1)^2 - 1 \] \ Add 1 to both sides of the equation to maintain equality: \ \[ (x - 1)^2 - 1 + y^2 + 4y = -1 + 1 \] \ Simplify: \ \[ (x - 1)^2 + y^2 + 4y = 0 \]
3Step 3: Complete the Square for y-terms
Next, complete the square for the \(y\) terms by taking the coefficient of \(y\), dividing by 2, and squaring it. \ The coefficient of \(y\) is 4. Dividing by 2 gives 2, and squaring gives 4. \ Rewrite the \(y\) terms: \ \[ (y^2 + 4y) \rightarrow (y + 2)^2 - 4 \] \ Add 4 to both sides of the equation: \ \[ (x - 1)^2 + (y + 2)^2 - 4 = 4 \] \ Simplify: \ \[ (x - 1)^2 + (y + 2)^2 = 4 \]
4Step 4: Identify the Circle Components
The equation \((x-1)^2 + (y+2)^2 = 4\) is in the standard form of a circle equation: \ \[(x-h)^2 + (y-k)^2 = r^2\] \ where \(h\) and \(k\) are the coordinates of the center, and \(r\) is the radius. \ \(h = 1\), \(k = -2\), and \(r^2 = 4\), so \(r = 2\). \ Thus, the center of the circle is \((1, -2)\) and the radius is 2.

Key Concepts

Completing the SquareCenter of a CircleRadius of a Circle
Completing the Square
Completing the square is a critical technique for transforming quadratic equations into a form that makes them easier to work with. This process is often used to rewrite the equation of a circle in its standard form:
  • First, focus on the terms containing the same variable.
  • For example, consider the expression \(x^2 - 2x\). To complete the square, identify the coefficient of \(x\), which is \(-2\).
  • Divide this coefficient by 2 to get \(-1\), then square it to obtain \(1\).
  • Now, rewrite \(x^2 - 2x\) as \((x - 1)^2 - 1\).
  • Similarly, tackle the \(y\)-terms by dividing the coefficient of \(y\), which is \(4\), by 2 to get \(2\), and squaring it to obtain \(4\).
  • Rewrite \(y^2 + 4y\) as \((y + 2)^2 - 4\).
By completing the square for both the \(x\) and \(y\) components, you convert the equation into a recognizably standard circle form. This simplifies identifying the circle's center and radius, which is crucial for geometry problems.
Center of a Circle
The center of a circle is a point from which all points on the circle are equidistant. In the standard circle equation \[(x-h)^2 + (y-k)^2 = r^2\],
  • the coordinates \((h, k)\) represent the center of the circle.
  • These coordinates are found by looking at the terms in the completed square forms, \((x-h)^2\) and \((y-k)^2\).
  • From these forms, \(h\) is the x-coordinate, and \(k\) is the y-coordinate of the center.
  • For example, in the equation \((x - 1)^2 + (y + 2)^2 = 4\), the center is \((1, -2)\).
  • It's important to remember that the values of \(h\) and \(k\) are often opposite in sign to those seen directly in the equation.
Finding the center is essential in problems where symmetry, rotations, or circle movements are involved. The center's location is always an anchor point in any circle-related geometry problem.
Radius of a Circle
The radius of a circle is one of the most fundamental measurements in geometry. It is the distance from the center of the circle to any point on its circumference.
  • In the standard equation \((x-h)^2 + (y-k)^2 = r^2\), \(r\) represents the radius.
  • This equation tells us that every point \((x, y)\) on the circle is \(r\) units away from the center \((h, k)\).
  • \(r^2\) is the term you see on the right side of the circle equation.
  • To find the radius from \(r^2\), take the square root of this value.
  • For instance, in an equation like \((x-1)^2 + (y+2)^2 = 4\), \(r^2 = 4\), so the radius \(r\) is 2.
Understanding the radius is crucial not only for determining the size of the circle but also for solving larger problems involving arcs, sectors, and real-world applications involving circular shapes.