Chapter 7: Problem 12
Complete the square in order to put the equation into standard form. Identify the center and the radius or explain why the equation does not represent a circle. $$ x^{2}+x+y^{2}-\frac{6}{5} y=1 $$
Short Answer
Expert verified
The center is \((-\frac{1}{2}, \frac{3}{5})\) and the radius is \(\frac{\sqrt{161}}{10}\).
Step by step solution
01
Group x and y Terms
The given equation is \( x^2 + x + y^2 - \frac{6}{5}y = 1 \). Start by rearranging the terms to group the \( x \) terms and \( y \) terms together: \( (x^2 + x) + (y^2 - \frac{6}{5}y) = 1 \).
02
Complete the Square for x
For the \( x \) terms \( x^2 + x \), take half of the coefficient of \( x \), which is \( \frac{1}{2} \), square it to get \( \left(\frac{1}{2}\right)^2 = \frac{1}{4} \), and add and subtract this inside the group: \( x^2 + x = (x^2 + x + \frac{1}{4} - \frac{1}{4}) \). Rewrite this as \( (x + \frac{1}{2})^2 - \frac{1}{4} \).
03
Complete the Square for y
For the \( y \) terms \( y^2 - \frac{6}{5}y \), take half of the coefficient of \( y \), which is \(-\frac{3}{5}\), square it to get \( \left(-\frac{3}{5}\right)^2 = \frac{9}{25} \), and add and subtract this inside the group: \( y^2 - \frac{6}{5}y = (y^2 - \frac{6}{5}y + \frac{9}{25} - \frac{9}{25}) \). Rewrite this as \( (y - \frac{3}{5})^2 - \frac{9}{25} \).
04
Combine Completed Squares
Substitute the completed square forms back into the equation: \[(x + \frac{1}{2})^2 - \frac{1}{4} + (y - \frac{3}{5})^2 - \frac{9}{25} = 1\].
05
Simplify Equation
Add \( \frac{1}{4} \) and \( \frac{9}{25} \) to both sides of the equation to balance it:\[(x + \frac{1}{2})^2 + (y - \frac{3}{5})^2 = 1 + \frac{1}{4} + \frac{9}{25}\].Convert the fractions and simplify the right side:\[1 = \frac{25}{25}, \frac{1}{4} = \frac{25}{100} = \frac{25}{100} \text{ which is equivalent to } \frac{25}{100}, \frac{9}{25} = \frac{36}{100} \].Thus the right side becomes: \[(x + \frac{1}{2})^2 + (y - \frac{3}{5})^2 = \frac{100}{100} + \frac{25}{100} + \frac{36}{100} = \frac{161}{100}\].
06
Identify Center and Radius
The standard form of a circle is \((x-h)^2 + (y-k)^2 = r^2 \), implying that the left-hand side of our equation matches the form, but the right side changed to a different value. So while it is a circle form equation after completion of square.Comparing, we identify the center as \( (h, k) = (-\frac{1}{2}, \frac{3}{5}) \) and the radius as \( r = \sqrt{\frac{161}{100}} = \frac{\sqrt{161}}{10} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Standard Form of a Circle
To understand how to determine the equation of a circle, we first need to grasp the concept of the "standard form of a circle". The standard form is written as \((x-h)^2 + (y-k)^2 = r^2\), where:
- \( (h, k) \) is the center of the circle.
- \( r \) is the radius of the circle.
Equation of a Circle
The equation \((x-h)^2 + (y-k)^2 = r^2\) describes a circle in a 2-dimensional plane. Here, \( (h, k) \) designates the fixed point or center from which every point on the circle is equidistant. The distance, denoted by \( r \), is the radius. Such an equation ensures that all the point coordinates \((x, y)\) at the calculated distance \( r \) from \( (h, k) \) make the circle complete.
For circles, an absence of an \( x \, y \) term (like +2xy in equations of ellipses or hyperbolas) keeps the shape uniform and round, maintaining perfect symmetry in all directions from the center. Identifying whether an equation is a true circle requires us to put it in its standard form and evaluate the symmetry of its terms.
For circles, an absence of an \( x \, y \) term (like +2xy in equations of ellipses or hyperbolas) keeps the shape uniform and round, maintaining perfect symmetry in all directions from the center. Identifying whether an equation is a true circle requires us to put it in its standard form and evaluate the symmetry of its terms.
Circle Center and Radius
Determining the center and radius of a circle from its equation is simple once the equation is in its standard form. In the equation \((x-h)^2 + (y-k)^2 = r^2\), the center \( (h, k) \) can be directly read as the values of \( h \) and \( k \) (keeping in mind the signs change). The radius \( r \) is the square root of the right side constant, \( r^2 \). For instance, if the transformed equation is \((x + \frac{1}{2})^2 + (y - \frac{3}{5})^2 = \frac{161}{100}\), comparing it to the standard form gives us:
- Center: \( (-\frac{1}{2}, \frac{3}{5}) \)
- Radius: \( r = \sqrt{\frac{161}{100}} = \frac{\sqrt{161}}{10} \)
Algebraic Transformation
Algebraic transformation, such as completing the square, is a technique used to convert quadratic equations into a form where properties of geometric shapes, like circles, become evident. Completing the square involves adjusting quadratic terms to create perfect square trinomials. This makes it easier to transform a general quadratic equation into its standard form.
For example, given an equation \(x^2 + x + y^2 - \frac{6}{5} y = 1\), group and complete the square for \( x \) and \( y \) terms separately:
For example, given an equation \(x^2 + x + y^2 - \frac{6}{5} y = 1\), group and complete the square for \( x \) and \( y \) terms separately:
- For \( x^2 + x \), add and subtract \( (\frac{1}{2})^2 \).
- For \( y^2 - \frac{6}{5} y \), add and subtract \( (\frac{3}{5})^2 \).