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 \((y + 2)^2 - 4\).
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.
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.
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