Chapter 11: Problem 80
Find the center and the radius of the circle with the given equation. [ 10.2] $$x^{2}+y^{2}-6 x+4 y=3$$
Short Answer
Expert verified
Center: (3, -2), Radius: 4
Step by step solution
01
Rearrange the equation
Start by rearranging the given equation into a form that groups the x and y terms together: \[ x^2 - 6x + y^2 + 4y = 3 \]
02
Complete the square for the x terms
To complete the square for the x terms, take the coefficient of x (-6), divide it by 2, and then square it: \[ (-6/2)^2 = 9 \]Add and subtract this value inside the equation: \[ x^2 - 6x + 9 - 9 + y^2 + 4y = 3 \]
03
Complete the square for the y terms
To complete the square for the y terms, take the coefficient of y (4), divide it by 2, and then square it: \[ (4/2)^2 = 4 \]Add and subtract this value inside the equation: \[ x^2 - 6x + 9 + y^2 + 4y + 4 = 3 + 9 + 4 \]
04
Rewrite the equation
Combine the completed squares and simplify: \[ (x-3)^2 + (y+2)^2 = 16 \]This is the standard form of the equation of a circle.
05
Identify the center and the radius
From the standard form \[ (x-h)^2 + (y-k)^2 = r^2 \], where \( h = 3 \) and \( k = -2 \), and \( r^2 = 16 \), the center is (3, -2) and the radius is 4.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Completing the Square
Completing the square is a method used to convert a quadratic equation into a perfect square trinomial. This is especially useful for finding the standard form of a circle's equation.
To complete the square for a variable, follow these steps:
To complete the square for a variable, follow these steps:
- Take the coefficient of the linear term (the x term or y term), divide it by 2, and then square it.
- Add and subtract this squared value in the equation. This helps in forming a perfect square trinomial.
- Coefficient of x is -6
- Divide by 2, we get -3
- Square it, which results in 9. Add and subtract 9 in the equation.
- Coefficient of y is 4
- Divide by 2, you get 2
- Square it to get 4. Add and subtract 4 in the equation.
Circle Center
Finding the center of a circle from its equation requires us to put it into standard form. The standard form of a circle's equation is \[ (x - h)^2 + (y - k)^2 = r^2 \] where \(h, k\) is the center, and \(r\) is the radius.
After completing the square, we get the equation: \[ (x - 3)^2 + (y + 2)^2 = 16 \] Here, we can directly identify the values:
After completing the square, we get the equation: \[ (x - 3)^2 + (y + 2)^2 = 16 \] Here, we can directly identify the values:
- The term \( (x - 3) \) indicates that \( h = 3 \).
- The term \( (y + 2) \) indicates that \( k = -2 \).
Circle Radius
Finding the radius of a circle from its equation is straightforward once it is in the standard form. Recall the standard form equation of a circle: \[ (x - h)^2 + (y - k)^2 = r^2 \].
In our example, the equation we derived is: \[ (x - 3)^2 + (y + 2)^2 = 16 \]. Here, \( r^2 = 16 \), which indicates:
In summary, converting the given equation to the standard form not only reveals the center but also the radius of the circle, providing a complete description of the circle's geometry.
In our example, the equation we derived is: \[ (x - 3)^2 + (y + 2)^2 = 16 \]. Here, \( r^2 = 16 \), which indicates:
- To find the radius \( r \), we take the square root of 16.
- The square root of 16 is 4.
In summary, converting the given equation to the standard form not only reveals the center but also the radius of the circle, providing a complete description of the circle's geometry.