Chapter 12: Problem 31
Write the center-radius form of the circle with the given equation. Give the center and radius. $$x^{2}+y^{2}+10 x-14 y-7=0$$
Short Answer
Expert verified
Center: (-5, 7); Radius: 9
Step by step solution
01
Group and move the constant
Group the x terms and y terms together and move the constant to the other side of the equation:\[ x^2 + 10x + y^2 - 14y = 7 \]
02
Complete the square for x
To complete the square for the x terms, take half of the coefficient of x, square it, and add it to both sides of the equation:Half of 10 is 5. Squaring 5 gives 25.\[ x^2 + 10x + 25 + y^2 - 14y = 7 + 25 \]This simplifies to:\[ (x + 5)^2 + y^2 - 14y = 32 \]
03
Complete the square for y
To complete the square for the y terms, take half of the coefficient of y, square it, and add it to both sides of the equation:Half of -14 is -7. Squaring -7 gives 49.\[ (x + 5)^2 + y^2 - 14y + 49 = 32 + 49 \]This simplifies to:\[ (x + 5)^2 + (y - 7)^2 = 81 \]
04
Write in center-radius form
The equation is now in the standard center-radius form of the circle equation which is \( (x - h)^2 + (y - k)^2 = r^2 \). Here, \( h = -5 \), \( k = 7 \) and \( r^2 = 81 \).
05
Identify center and radius
Extract the center and radius from the equation:Center: \( (-5, 7) \)Radius: \( \sqrt{81} = 9 \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Center-Radius Form
The center-radius form of a circle is a very useful way to represent the equation of a circle. This form is given by\((x - h)^2 + (y - k)^2 = r^2\), where \( (h, k) \) represents the coordinates of the center of the circle and \( r \) represents the radius.
This format makes it easy to identify the key attributes of a circle.
When a circle's equation is in this form, you immediately know:
This format makes it easy to identify the key attributes of a circle.
When a circle's equation is in this form, you immediately know:
- The center, by looking at \(h\) and \(k\)
- The radius, by taking the square root of \(r^2\)
Completing the Square
Completing the square is a method used to convert a quadratic equation into a perfect square trinomial. This technique is very useful in circle equations as it helps to bring the equation into the center-radius form.
To complete the square, follow these steps:
\((x^2 + 10x + 25) + (y^2 - 14y + 49) = 7 + 25 + 49\)
which simplifies to \((x + 5)^2 + (y - 7)^2 = 81\). This process transforms the equation into a form that easily reveals the circle's center and radius.
To complete the square, follow these steps:
- Take the quadratic terms (e.g., \(x^2\) and \(y^2\))
- Group them with their respective linear terms (e.g., \(10x\) and \(-14y\))
- Find half of the linear coefficient, square it, and add it to both sides of the equation
\((x^2 + 10x + 25) + (y^2 - 14y + 49) = 7 + 25 + 49\)
which simplifies to \((x + 5)^2 + (y - 7)^2 = 81\). This process transforms the equation into a form that easily reveals the circle's center and radius.
Circle Geometry
Understanding circle geometry is essential when dealing with circle equations. A circle is defined as the set of all points that are equidistant from a central point. This central point is called the center, and the distance from the center to any point on the circle is the radius.
There are a few important properties:
There are a few important properties:
- All radii in a circle are equal
- The center of the circle is a fixed point
- The radius is the distance from this center to any point on the circle
Standard Form of a Circle
The standard form of a circle is another way to write a circle's equation, and it leads into the center-radius form. The standard form is often written as \(x^2 + y^2 + Dx + Ey + F = 0\). To convert this into the center-radius form, you use the method of completing the square.
Starting with our standard form equation:
\( x^2 + y^2 + 10x - 14y - 7 = 0\), we complete the square for both x and y terms as shown in the provided solution.
After completing the square, we convert it to the center-radius form of \((x + 5)^2 + (y - 7)^2 = 81\). This conversion is vital because it allows us to easily extract the circle's center and radius, making both geometric and algebraic interpretations straightforward.
Starting with our standard form equation:
\( x^2 + y^2 + 10x - 14y - 7 = 0\), we complete the square for both x and y terms as shown in the provided solution.
After completing the square, we convert it to the center-radius form of \((x + 5)^2 + (y - 7)^2 = 81\). This conversion is vital because it allows us to easily extract the circle's center and radius, making both geometric and algebraic interpretations straightforward.