Chapter 10: Problem 47
Find the center and radius of the circle with equation \(x^{2}+y^{2}-10 x+2 y+\) \(22=0 .\) Then graph the circle.
Short Answer
Expert verified
The circle has center (5, -1) and radius 2.
Step by step solution
01
Write the original equation
Let's start by citing the original equation of the circle: \ \(x^{2}+y^{2}-10x+2y+22=0\).
02
Rearrange the equation
Move the constant term to the right-hand side by subtracting 22 from both sides. This gives: \ \(x^{2} + y^{2} - 10x + 2y = -22\).
03
Complete the square for the x terms
Identify the coefficient of \(x\), which is \(-10\). Take half of it, \(\frac{-10}{2} = -5\), and square it, \((-5)^{2} = 25\). Add and subtract 25 to complete the square: \ \((x^{2} - 10x + 25)\).
04
Complete the square for the y terms
Identify the coefficient of \(y\), which is \(2\). Take half of it, \(\frac{2}{2} = 1\), and square it, \(1^{2} = 1\). Add and subtract 1 to complete the square: \ \((y^{2} + 2y + 1)\).
05
Rewrite the equation in completed square form
Incorporate the completed squares: \ \((x-5)^2 + (y+1)^2 = -22 + 25 + 1\). This simplifies to: \ \((x-5)^2 + (y+1)^2 = 4\).
06
Identify the center of the circle
The equation \((x-5)^2 + (y+1)^2 = 4\) is in the standard form \ \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center. Thus, the center is \((5, -1)\).
07
Identify the radius of the circle
In the equation \((x-5)^2 + (y+1)^2 = 4\), we see that \(r^2 = 4\). Therefore, the radius \(r\) is \(\sqrt{4} = 2\).
08
Graph the circle
To graph the circle, plot the center at \((5, -1)\). Then draw a circle around this center with a radius of 2, ensuring all points are equidistant from the center by 2 units.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Center of a Circle
The center of a circle is a critical point that defines the circle's position in the coordinate plane. For the equation of a circle \((x - h)^2 + (y - k)^2 = r^2\), the center is represented by the coordinates \((h, k)\). This is where the circle "sits", and all the points on the circle are equidistant from this point.
- Understanding Coordinate Shift: When looking at the terms \((x - h)\) and \((y - k)\), subtracting a number directly points out the shift from the origin to the new center \((h, k)\).
- Example: In the problem solved, the equation \( (x-5)^2 + (y+1)^2 = 4 \) tells us the center is at \((5, -1)\).
Radius of a Circle
The radius of a circle is the distance from the center to any point on the circle's edge. In the equation \((x - h)^2 + (y - k)^2 = r^2\), the radius \(r\) is found by taking the square root of the value on the right side.
- Mathematical Definition: It is mathematically expressed as \(r = \sqrt{r^2}\).
- Example: In the given exercise, the right-hand side of the transformed equation is 4, meaning \(r^2 = 4\). This implies that the radius \(r = 2\).
Completing the Square
Completing the square is a technique used to transform a quadratic expression into a perfect square trinomial. This technique is indispensable in forming the standard equation of a circle.
- Process: It involves taking the coefficient of \(x\) or \(y\), dividing it by 2, squaring it, and using this value to adjust the equation.
- Example: For the \(x\)-terms \(x^2 - 10x\), half of -10 is -5, squaring gives 25. Therefore, the expression becomes \( (x-5)^2\).
Standard Form of a Circle Equation
The standard form of a circle equation is \((x-h)^2 + (y-k)^2 = r^2\), which provides a clear path to identifying the circle's key features like center and radius.
- Importance: This form makes it easy to quickly see the circle's center and radius.
- Transformation: Originating from a general quadratic form, the process of rearranging and completing the square converts it to this standard form.
- Visual Clarity: When plotting, knowing the center \((h, k)\) and radius \(r\) helps you draft a perfect circle in any graphical representation.