Chapter 2: Problem 67
\(65-72\) . Show that the equation represents a circle, and find the center and radius of the circle. $$ x^{2}+y^{2}-4 x+10 y+13=0 $$
Short Answer
Expert verified
The circle's center is (2, -5) and the radius is 4.
Step by step solution
01
Identify the General Form
The given equation is \(x^{2} + y^{2} - 4x + 10y + 13 = 0\). Identify its terms: quadratic terms are \(x^{2}\) and \(y^{2}\); linear terms are \(-4x\) and \(10y\); and the constant is \(13\).
02
Group and Complete the Square for x
Focus on the terms involving \(x\): \(x^{2} - 4x\). To complete the square, take the coefficient of \(x\), which is \(-4\), divide by 2 to get \(-2\), and square it to add \(4\). This transforms \(x^{2} - 4x\) into \((x - 2)^{2}\). Subtract \(4\) to maintain equality.
03
Group and Complete the Square for y
Now focus on the terms involving \(y\): \(y^{2} + 10y\). Take the coefficient of \(y\), which is \(10\), divide by 2 to get \(5\), and square it to add \(25\). This transforms \(y^{2} + 10y\) into \((y + 5)^{2}\). Subtract \(25\) to keep the equation balanced.
04
Rewrite the Equation in Circle Form
Update the equation with the completed squares: \((x - 2)^{2} - 4 + (y + 5)^{2} - 25 + 13 = 0\). Simplify the constant terms: \(-4 - 25 + 13 = -16\). Hence, the equation becomes \((x - 2)^{2} + (y + 5)^{2} = 16\).
05
Identify Center and Radius
The circle's equation is \((x - h)^{2} + (y - k)^{2} = r^{2}\). From \((x - 2)^{2} + (y + 5)^{2} = 16\), the center \((h, k)\) is \((2, -5)\) and the radius \(r\) is \(\sqrt{16} = 4\).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Completing the Square
Completing the square is a key mathematical method that helps in rewriting quadratic expressions to reveal their characteristics more clearly. Imagine you have any terms such as \(x^2 - 4x\). Here, our goal is to transform these terms into a perfect square trinomial. Why do we do this? Because perfect squares are easier to work with and give us valuable insights.
To complete the square for a term like \(x^2 - 4x\), the method is simple and systematic:
To complete the square for a term like \(x^2 - 4x\), the method is simple and systematic:
- Take the coefficient of \(x\), which is \(-4\).
- Divide it by 2, yielding \(-2\).
- Square this result to get \(4\).
Center of a Circle
The center of a circle in coordinate geometry is vital as it serves as the anchor point from which every point on the circle maintains a constant distance (the radius). When dealing with circle equations in the format \((x - h)^2 + (y - k)^2 = r^2\), the coordinates \((h, k)\) represent the circle's center.
In our specific example, after completing the square, we derived the equation \((x - 2)^2 + (y + 5)^2 = 16\). From this, we can see the values \(h = 2\) and \(k = -5\). Hence, the center of this circle is at \((2, -5)\).
It is important to remember that the expressions \(h\) and \(-k\) in the equation tell us exactly how much we "shift" the basic circle \((x)^2 + (y)^2 = r^2\) from the origin, making graphing and understanding its position straightforward.
In our specific example, after completing the square, we derived the equation \((x - 2)^2 + (y + 5)^2 = 16\). From this, we can see the values \(h = 2\) and \(k = -5\). Hence, the center of this circle is at \((2, -5)\).
It is important to remember that the expressions \(h\) and \(-k\) in the equation tell us exactly how much we "shift" the basic circle \((x)^2 + (y)^2 = r^2\) from the origin, making graphing and understanding its position straightforward.
Radius of a Circle
Finding the radius of a circle using its equation is straightforward yet crucial.
The equation of a circle often appears in the form \((x - h)^2 + (y - k)^2 = r^2\), where \(r^2\) is the radius squared. Knowing \(r^2\) allows us to easily calculate the radius, \(r\), by taking the square root.
In our processed form of the circle equation \((x - 2)^2 + (y + 5)^2 = 16\), the term \(16\) represents \(r^2\). Therefore, the radius \(r\) is the square root: \(\sqrt{16} = 4\).
This tells us that every point on the circle sits exactly 4 units away from its center. Understanding the radius is crucial not just for graphing, but also for solving real-life problems involving circles where knowing a fixed distance from the center is important.
The equation of a circle often appears in the form \((x - h)^2 + (y - k)^2 = r^2\), where \(r^2\) is the radius squared. Knowing \(r^2\) allows us to easily calculate the radius, \(r\), by taking the square root.
In our processed form of the circle equation \((x - 2)^2 + (y + 5)^2 = 16\), the term \(16\) represents \(r^2\). Therefore, the radius \(r\) is the square root: \(\sqrt{16} = 4\).
This tells us that every point on the circle sits exactly 4 units away from its center. Understanding the radius is crucial not just for graphing, but also for solving real-life problems involving circles where knowing a fixed distance from the center is important.