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Magazine Chapter 1: Problem 14
Complete the square in \(x\) and \(y\) to find the center and the radius of the given circle. \(\frac{1}{2} x^{2}+\frac{1}{2} y^{2}+\frac{5}{2} x+10 y+5=0\)
Short Answer
Expert verified
Center: \((-\frac{5}{2}, -10)\), Radius: \(\frac{\sqrt{385}}{2}\).
Step by step solution
01
Rearrange the Equation
First, multiply the entire equation by 2 to eliminate the fractions: \[ x^2 + y^2 + 5x + 20y + 10 = 0 \] Then, move the constant term to the right side of the equation: \[ x^2 + y^2 + 5x + 20y = -10 \]
02
Complete the square for x-terms
Start with the x terms: \[ x^2 + 5x \] To complete the square, take the coefficient of x, which is 5, divide it by 2 and square it: \[ \left(\frac{5}{2}\right)^2 = \frac{25}{4} \] Add and subtract \(\frac{25}{4}\) inside the equation: \[ x^2 + 5x + \frac{25}{4} - \frac{25}{4} \] The first three terms form a perfect square trinomial: \[ (x + \frac{5}{2})^2 - \frac{25}{4} \]
03
Complete the square for y-terms
Now, focus on the y terms: \[ y^2 + 20y \] Take the coefficient of y, which is 20, divide it by 2, and square it: \[ \left(\frac{20}{2}\right)^2 = 100 \] Add and subtract 100: \[ y^2 + 20y + 100 - 100 \] This becomes a perfect square trinomial:\[ (y + 10)^2 - 100 \]
04
Substitute back into the equation
Substitute the completed squares back into the equation: \[ (x + \frac{5}{2})^2 - \frac{25}{4} + (y + 10)^2 - 100 = -10 \] Combine constants on the right-hand side: \[ (x + \frac{5}{2})^2 + (y + 10)^2 = -10 + \frac{25}{4} + 100 \] Convert \(-10\) to quarter form: \[ -10 = -\frac{40}{4} \] Combine all fractions: \[ (x + \frac{5}{2})^2 + (y + 10)^2 = \frac{25}{4} + \frac{400}{4} - \frac{40}{4} \] Simplify: \[ = \frac{385}{4} \]
05
Determine the center and radius
The equation \[ (x + \frac{5}{2})^2 + (y + 10)^2 = \frac{385}{4} \] is in the standard form of a circle equation: \[ (x - h)^2 + (y - k)^2 = r^2 \] Here, \( h = -\frac{5}{2} \), \( k = -10 \), and \( r^2 = \frac{385}{4} \), so the radius \( r \) is \( r = \sqrt{\frac{385}{4}} = \frac{\sqrt{385}}{2} \). Thus, the center of the circle is \( \left(-\frac{5}{2}, -10\right) \) and the radius is \( \frac{\sqrt{385}}{2} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Circle Equations
The equation of a circle in mathematics is an essential concept that helps us to define the set of all points that are equidistant from a given central point in a plane. The standard form of a circle equation is: \[ (x-h)^2 + (y-k)^2 = r^2 \]
- Here, \( (h, k) \) represents the center of the circle.
- \( r \) signifies the radius, which is the distance between the center and any point on the circle.
Center and Radius of a Circle
Understanding the center and the radius from the circle equation allows us to determine the circle's size and location on a coordinate plane. Let's recap the equation we dealt with:\[ (x + \frac{5}{2})^2 + (y + 10)^2 = \frac{385}{4} \]
- The expression \((x + \frac{5}{2})\) implies the x-coordinate of the center is \(-\frac{5}{2}\). Likewise, \((y + 10)\) tells us the y-coordinate is \(-10\).
- This places the circle's center at \((-\frac{5}{2}, -10)\).
- The radius squared, \(r^2\), is \(\frac{385}{4}\), so the radius \(r\) is \(\frac{\sqrt{385}}{2}\).
Perfect Square Trinomial
Completing the square is a mathematical technique used to rewrite quadratic equations and identify forms such as perfect square trinomials. A trinomial is considered a perfect square if it can be expressed as the square of a binomial. For example, \[ a^2 + 2ab + b^2 = (a+b)^2 \].
In the given problem, to form a perfect square trinomial from the x terms \(x^2 + 5x\), we:
In the given problem, to form a perfect square trinomial from the x terms \(x^2 + 5x\), we:
- Took \(5\), divided by \(2\) (giving \(\frac{5}{2}\)), and squared it to make \(\frac{25}{4}\).
- This reformulated the expression to: \((x + \frac{5}{2})^2 - \frac{25}{4}\).
- Divide \(20\) by \(2\) and square to obtain \(100\).
- Transforming this segment to: \((y + 10)^2 - 100\).
